{"id":62998,"date":"2023-08-07T22:41:00","date_gmt":"2023-08-07T17:11:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=62998"},"modified":"2023-12-08T10:12:47","modified_gmt":"2023-12-08T04:42:47","slug":"rd-sharma-solutions-class-11-maths-chapter-18","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/","title":{"rendered":"RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem (Updated For 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-119499\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 18 \" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 18<\/strong> &#8211; Binomial Theorem:\u00a0<span style=\"font-weight: 400;\">It is RD Sharma Solutions Class 11 Maths Chapter 18 Binomial Theorem in which <\/span><span style=\"font-weight: 400;\">all the parts of the blog are associated with the fundamental aspects of Binomial Theorem<\/span><span style=\"font-weight: 400;\"> like the benefits of following <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a> Chapter 18 Maths Solutions along with exercise-wise explanations.\u00a0<\/span><\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e73df5889f8\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e73df5889f8\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#download-rd-sharma-solutions-class-11-maths-chapter-18-%e2%80%93-binomial-theorem-pdf\" title=\"Download RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem PDF\">Download RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#exercise-wise-rd-sharma-solutions-class-11-maths-chapter-18\" title=\"Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 18\">Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 18<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#access-rd-sharma-solutions-class-11-maths-chapter-18\" title=\"Access RD Sharma Solutions Class 11 Maths Chapter 18\u00a0\">Access RD Sharma Solutions Class 11 Maths Chapter 18\u00a0<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#detailed-exercise-wise-explanation-with-listing-of-important-topics-in-the-exercise\" title=\"Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise\">Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#rd-sharma-class-11-chapter-18-exercise-18a\" title=\"RD Sharma Class 11 Chapter 18 Exercise 18A\">RD Sharma Class 11 Chapter 18 Exercise 18A<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#rd-sharma-class-11-chapter-18-exercise-18b\" title=\"RD Sharma Class 11 Chapter 18 Exercise 18B\">RD Sharma Class 11 Chapter 18 Exercise 18B<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#important-concepts-from-rd-sharma-solutions-class-11-maths-chapter-18\" title=\"Important concepts from RD Sharma Solutions Class 11 Maths Chapter 18\">Important concepts from RD Sharma Solutions Class 11 Maths Chapter 18<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#rd-sharma-solutions-class-11-maths-chapter-18\" title=\"RD Sharma Solutions Class 11 Maths Chapter 18\">RD Sharma Solutions Class 11 Maths Chapter 18<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#access-other-important-chapters-of-rd-sharma-solutions-class-11-maths\" title=\"Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0\">Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#faqs-on-rd-sharma-solutions-class-11-maths-chapter-18\" title=\"FAQs on RD Sharma Solutions Class 11 Maths Chapter 18\">FAQs on RD Sharma Solutions Class 11 Maths Chapter 18<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-18\" title=\"From where can I download the PDF of\u00a0RD Sharma Solutions Class 11 Maths Chapter 18?\">From where can I download the PDF of\u00a0RD Sharma Solutions Class 11 Maths Chapter 18?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-18\" title=\"How much does it cost to download the PDF of\u00a0RD Sharma Solutions Class 11 Maths Chapter 18?\">How much does it cost to download the PDF of\u00a0RD Sharma Solutions Class 11 Maths Chapter 18?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/#can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-18-pdf-offline\" title=\"Can I access the\u00a0RD Sharma Solutions Class 11 Maths Chapter 18 PDF offline?\">Can I access the\u00a0RD Sharma Solutions Class 11 Maths Chapter 18 PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-18-%e2%80%93-binomial-theorem-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 18<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/08\/RD-Sharma-Solutions-Class-11-Maths-Chapter-18-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"exercise-wise-rd-sharma-solutions-class-11-maths-chapter-18\"><\/span>Exercise-Wise RD Sharma Solutions Class 11 Maths Chapter 18<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-1\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Exercise 18.1<\/a><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 100%;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18-exercise-18-2\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Exercise 18.2<\/a><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h3><span class=\"ez-toc-section\" id=\"access-rd-sharma-solutions-class-11-maths-chapter-18\"><\/span><span style=\"font-size: 30px; background-color: initial;\">Access RD Sharma Solutions Class 11 Maths Chapter 18\u00a0<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Using the binomial theorem, write down the expressions of the following:<\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-1.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 1\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i) (2x + 3y)<sup>\u00a05<\/sup><\/strong><\/p>\n<p>Let us solve the given expression<\/p>\n<p>(2x + 3y)<sup>\u00a05<\/sup>\u00a0=\u00a0<sup>5<\/sup>C<sub>0<\/sub>\u00a0(2x)<sup>5<\/sup>\u00a0(3y)<sup>0<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>1<\/sub>\u00a0(2x)<sup>4<\/sup>\u00a0(3y)<sup>1<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(2x)<sup>3<\/sup>\u00a0(3y)<sup>2<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(2x)<sup>2<\/sup>\u00a0(3y)<sup>3<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(2x)<sup>1<\/sup>\u00a0(3y)<sup>4<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(2x)<sup>0<\/sup>\u00a0(3y)<sup>5<\/sup><\/p>\n<p>= 32x<sup>5<\/sup>\u00a0+ 5 (16x<sup>4<\/sup>) (3y) + 10 (8x<sup>3<\/sup>) (9y)<sup>2<\/sup>\u00a0+ 10 (4x)<sup>2<\/sup>\u00a0(27y)<sup>3<\/sup>\u00a0+ 5 (2x) (81y<sup>4<\/sup>) + 243y<sup>5<\/sup><\/p>\n<p><em>= 32x<sup>5<\/sup>\u00a0+ 240x<sup>4<\/sup>y + 720x<sup>3<\/sup>y<sup>2<\/sup>\u00a0+ 1080x<sup>2<\/sup>y<sup>3<\/sup>\u00a0+ 810xy<sup>4<\/sup>\u00a0+ 243y<sup>5<\/sup><\/em><\/p>\n<p><strong>(ii) (2x \u2013 3y)<sup>\u00a04<\/sup><\/strong><\/p>\n<p>Let us solve the given expression<\/p>\n<p>(2x \u2013 3y)<sup>\u00a04<\/sup>\u00a0=\u00a0<sup>4<\/sup>C<sub>0<\/sub>\u00a0(2x)<sup>4<\/sup>\u00a0(3y)<sup>0<\/sup>\u00a0\u2013\u00a0<sup>4<\/sup>C<sub>1<\/sub>\u00a0(2x)<sup>3<\/sup>\u00a0(3y)<sup>1<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>2<\/sub>\u00a0(2x)<sup>2<\/sup>\u00a0(3y)<sup>2<\/sup>\u00a0\u2013\u00a0<sup>4<\/sup>C<sub>3<\/sub>\u00a0(2x)<sup>1<\/sup>\u00a0(3y)<sup>3<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>4<\/sub>\u00a0(2x)<sup>0<\/sup>\u00a0(3y)<sup>4<\/sup><\/p>\n<p>= 16x<sup>4<\/sup>\u00a0\u2013 4 (8x<sup>3<\/sup>) (3y) + 6 (4x<sup>2<\/sup>) (9y<sup>2<\/sup>) \u2013 4 (2x) (27y<sup>3<\/sup>) + 81y<sup>4<\/sup><\/p>\n<p><em>= 16x<sup>4<\/sup>\u00a0\u2013 96x<sup>3<\/sup>y + 216x<sup>2<\/sup>y<sup>2<\/sup>\u00a0\u2013 216xy<sup>3<\/sup>\u00a0+ 81y<sup>4<\/sup><\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-2.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 2\" \/><\/p>\n<p><strong>(iv) (1 \u2013 3x)<sup>\u00a07<\/sup><\/strong><\/p>\n<p>Let us solve the given expression<\/p>\n<p>(1 \u2013 3x)<sup>\u00a07<\/sup>\u00a0=\u00a0<sup>7<\/sup>C<sub>0<\/sub>\u00a0(3x)<sup>0<\/sup>\u00a0\u2013\u00a0<sup>7<\/sup>C<sub>1<\/sub>\u00a0(3x)<sup>1<\/sup>\u00a0+\u00a0<sup>7<\/sup>C<sub>2<\/sub>\u00a0(3x)<sup>2<\/sup>\u00a0\u2013\u00a0<sup>7<\/sup>C<sub>3<\/sub>\u00a0(3x)<sup>3<\/sup>\u00a0+\u00a0<sup>7<\/sup>C<sub>4<\/sub>\u00a0(3x)<sup>4<\/sup>\u00a0\u2013\u00a0\u00a0<sup>7<\/sup>C<sub>5<\/sub>\u00a0(3x)<sup>5<\/sup>\u00a0+\u00a0<sup>7<\/sup>C<sub>6<\/sub>\u00a0(3x)<sup>6<\/sup>\u00a0\u2013\u00a0<sup>7<\/sup>C<sub>7<\/sub>\u00a0(3x)<sup>7<\/sup><\/p>\n<p>= 1 \u2013 7 (3x) + 21 (9x)<sup>2<\/sup>\u00a0\u2013 35 (27x<sup>3<\/sup>) + 35 (81x<sup>4<\/sup>) \u2013 21 (243x<sup>5<\/sup>) + 7 (729x<sup>6<\/sup>) \u2013 2187(x<sup>7<\/sup>)<\/p>\n<p><em>= 1 \u2013 21x + 189x<sup>2<\/sup>\u00a0\u2013 945x<sup>3<\/sup>\u00a0+ 2835x<sup>4<\/sup>\u00a0\u2013 5103x<sup>5<\/sup>\u00a0+ 5103x<sup>6<\/sup>\u00a0\u2013 2187x<sup>7<\/sup><\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-3.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 3\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-4.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 4\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-5.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 5\" \/><\/p>\n<p><strong>(viii) (1 + 2x \u2013 3x<sup>2<\/sup>)<sup>5<\/sup><\/strong><\/p>\n<p>Let us solve the given expression<\/p>\n<p>Let us consider (1 + 2x) and 3x<sup>2<\/sup>\u00a0as two different entities and apply the binomial theorem.<\/p>\n<p>(1 + 2x \u2013 3x<sup>2<\/sup>)<sup>5<\/sup>\u00a0=\u00a0<sup>5<\/sup>C<sub>0<\/sub>\u00a0(1 + 2x)<sup>5<\/sup>\u00a0(3x<sup>2<\/sup>)<sup>0<\/sup>\u00a0\u2013\u00a0<sup>5<\/sup>C<sub>1<\/sub>\u00a0(1 + 2x)<sup>4<\/sup>\u00a0(3x<sup>2<\/sup>)<sup>1<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(1 + 2x)<sup>3<\/sup>\u00a0(3x<sup>2<\/sup>)<sup>2<\/sup>\u00a0\u2013\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(1 + 2x)<sup>2<\/sup>\u00a0(3x<sup>2<\/sup>)<sup>3<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(1 + 2x)<sup>1<\/sup>\u00a0(3x<sup>2<\/sup>)<sup>4<\/sup>\u00a0\u2013\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(1 + 2x)<sup>0<\/sup>\u00a0(3x<sup>2<\/sup>)<sup>5<\/sup><\/p>\n<p>= (1 + 2x)<sup>5<\/sup>\u00a0\u2013 5(1 + 2x)<sup>4<\/sup>\u00a0(3x<sup>2<\/sup>) + 10 (1 + 2x)<sup>3<\/sup>\u00a0(9x<sup>4<\/sup>) \u2013 10 (1 + 2x)<sup>2<\/sup>\u00a0(27x<sup>6<\/sup>) + 5 (1 + 2x) (81x<sup>8<\/sup>) \u2013 243x<sup>10<\/sup><\/p>\n<p>=\u00a0<sup>5<\/sup>C<sub>0<\/sub>\u00a0(2x)<sup>0<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>1<\/sub>\u00a0(2x)<sup>1<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(2x)<sup>2<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(2x)<sup>3<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(2x)<sup>4<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(2x)<sup>5\u00a0<\/sup>\u2013 15x<sup>2<\/sup>\u00a0[<sup>4<\/sup>C<sub>0<\/sub>\u00a0(2x)<sup>0<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>1<\/sub>\u00a0(2x)<sup>1<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>2<\/sub>\u00a0(2x)<sup>2<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>3<\/sub>\u00a0(2x)<sup>3<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>4<\/sub>\u00a0(2x)<sup>4<\/sup>] + 90x<sup>4<\/sup>\u00a0[1 + 8x<sup>3<\/sup>\u00a0+ 6x + 12x<sup>2<\/sup>] \u2013 270x<sup>6<\/sup>(1 + 4x<sup>2<\/sup>\u00a0+ 4x) + 405x<sup>8<\/sup>\u00a0+ 810x<sup>9<\/sup>\u00a0\u2013 243x<sup>10<\/sup><\/p>\n<p>= 1 + 10x + 40x<sup>2<\/sup>\u00a0+ 80x<sup>3<\/sup>\u00a0+ 80x<sup>4<\/sup>\u00a0+ 32x<sup>5<\/sup>\u00a0\u2013 15x<sup>2<\/sup>\u00a0\u2013 120x<sup>3<\/sup>\u00a0\u2013 360<sup>4<\/sup>\u00a0\u2013 480x<sup>5<\/sup>\u00a0\u2013 240x<sup>6<\/sup>\u00a0+ 90x<sup>4<\/sup>\u00a0+ 720x<sup>7<\/sup>\u00a0+ 540x<sup>5<\/sup>\u00a0+ 1080x<sup>6<\/sup>\u00a0\u2013 270x<sup>6<\/sup>\u00a0\u2013 1080x<sup>8<\/sup>\u00a0\u2013 1080x<sup>7<\/sup>\u00a0+ 405x<sup>8<\/sup>\u00a0+ 810x<sup>9<\/sup>\u00a0\u2013 243x<sup>10<\/sup><\/p>\n<p><em>= 1 + 10x + 25x<sup>2<\/sup>\u00a0\u2013 40x<sup>3<\/sup>\u00a0\u2013 190x<sup>4<\/sup>\u00a0+ 92x<sup>5<\/sup>\u00a0+ 570x<sup>6<\/sup>\u00a0\u2013 360x<sup>7<\/sup>\u00a0\u2013 675x<sup>8<\/sup>\u00a0+ 810x<sup>9<\/sup>\u00a0\u2013 243x<sup>10<\/sup><\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-6.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 6\" \/><\/p>\n<p><strong>(x)\u00a0<\/strong>(1 \u2013 2x + 3x<sup>2<\/sup>)<sup>3<\/sup><\/p>\n<p>Let us solve the given expression<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-7.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 7\" \/><\/p>\n<p><strong>2. Evaluate the following:<\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-8.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 8\" \/><\/strong><\/p>\n<p><strong><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-9.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 9\" \/><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-10.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 10\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-11.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 11\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-12.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 12\" \/><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-13.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 13\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= 2 [<sup>5<\/sup>C<sub>0<\/sub>\u00a0(2<strong>\u221a<\/strong>x)<sup>0<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(2<strong>\u221a<\/strong>x)<sup>2<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(2<strong>\u221a<\/strong>x)<sup>4<\/sup>]<\/p>\n<p>= 2 [1 + 10 (4x) + 5 (16x<sup>2<\/sup>)]<\/p>\n<p><em>= 2 [1 + 40x + 80x<sup>2<\/sup>]<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-14.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 14\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= 2 [<sup>6<\/sup>C<sub>0<\/sub>\u00a0(<strong>\u221a<\/strong>2)<sup>6<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>2<\/sub>\u00a0(<strong>\u221a<\/strong>2)<sup>4<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>4<\/sub>\u00a0(<strong>\u221a<\/strong>2)<sup>2<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>6<\/sub>\u00a0(<strong>\u221a<\/strong>2)<sup>0<\/sup>]<\/p>\n<p>= 2 [8 + 15 (4) + 15 (2) + 1]<\/p>\n<p>= 2 [99]<\/p>\n<p><em>= 198<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-15.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 15\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= 2 [<sup>5<\/sup>C<sub>1<\/sub>\u00a0(3<sup>4<\/sup>) (<strong>\u221a<\/strong>2)<sup>1<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(3<sup>2<\/sup>) (<strong>\u221a<\/strong>2)<sup>3<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(3<sup>0<\/sup>) (<strong>\u221a<\/strong>2)<sup>5<\/sup>]<\/p>\n<p>= 2 [5 (81) (<strong>\u221a<\/strong>2) + 10 (9) (2<strong>\u221a<\/strong>2) + 4<strong>\u221a<\/strong>2]<\/p>\n<p>= 2<strong>\u221a<\/strong>2 (405 + 180 + 4)<\/p>\n<p><em>= 1178<strong>\u221a<\/strong>2<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-16.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 16\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= 2 [<sup>7<\/sup>C<sub>0<\/sub>\u00a0(2<sup>7<\/sup>) (<strong>\u221a<\/strong>3)<sup>0<\/sup>\u00a0+\u00a0<sup>7<\/sup>C<sub>2<\/sub>\u00a0(2<sup>5<\/sup>) (<strong>\u221a<\/strong>3)<sup>2<\/sup>\u00a0+\u00a0<sup>7<\/sup>C<sub>4<\/sub>\u00a0(2<sup>3<\/sup>) (<strong>\u221a<\/strong>3)<sup>4<\/sup>\u00a0+\u00a0<sup>7<\/sup>C<sub>6<\/sub>\u00a0(2<sup>1<\/sup>) (<strong>\u221a<\/strong>3)<sup>6<\/sup>]<\/p>\n<p>= 2 [128 + 21 (32)(3) + 35(8)(9) + 7(2)(27)]<\/p>\n<p>= 2 [128 + 2016 + 2520 + 378]<\/p>\n<p>= 2 [5042]<\/p>\n<p><em>= 10084<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-17.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 17\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= 2 [<sup>5<\/sup>C<sub>1<\/sub>\u00a0(<strong>\u221a<\/strong>3)<sup>4<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(<strong>\u221a<\/strong>3)<sup>2<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(<strong>\u221a<\/strong>3)<sup>0<\/sup>]<\/p>\n<p>= 2 [5 (9) + 10 (3) + 1]<\/p>\n<p>= 2 [76]<\/p>\n<p><em>= 152<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-18.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 18\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= (1 \u2013 0.01)<sup>5<\/sup>\u00a0+ (1 + 0.01)<sup>5<\/sup><\/p>\n<p>= 2 [<sup>5<\/sup>C<sub>0<\/sub>\u00a0(0.01)<sup>0<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(0.01)<sup>2<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(0.01)<sup>4<\/sup>]<\/p>\n<p>= 2 [1 + 10 (0.0001) + 5 (0.00000001)]<\/p>\n<p>= 2 [1.00100005]<\/p>\n<p><em>= 2.0020001<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-19.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 19\" \/><\/p>\n<p>Let us solve the given expression<\/p>\n<p>= 2 [<sup>6<\/sup>C<sub>1<\/sub>\u00a0(<strong>\u221a<\/strong>3)<sup>5<\/sup>\u00a0(<strong>\u221a<\/strong>2)<sup>1<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>3<\/sub>\u00a0(<strong>\u221a<\/strong>3)<sup>3<\/sup>\u00a0(<strong>\u221a<\/strong>2)<sup>3<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>5<\/sub>\u00a0(<strong>\u221a<\/strong>3)<sup>1<\/sup>\u00a0(<strong>\u221a<\/strong>2)<sup>5<\/sup>]<\/p>\n<p>= 2 [6 (9<strong>\u221a<\/strong>3) (<strong>\u221a<\/strong>2) + 20 (3<strong>\u221a<\/strong>3) (2<strong>\u221a<\/strong>2) + 6 (<strong>\u221a<\/strong>3) (4<strong>\u221a<\/strong>2)]<\/p>\n<p>= 2 [<strong>\u221a<\/strong>6 (54 + 120 + 24)]<\/p>\n<p><em>= 396\u00a0<strong>\u221a<\/strong>6<\/em><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-20.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 20\" \/><\/p>\n<p>= 2 [a<sup>8<\/sup>\u00a0+ 6a<sup>6<\/sup>\u00a0\u2013 6a<sup>4<\/sup>\u00a0+ a<sup>4<\/sup>\u00a0+ 1 \u2013 2a<sup>2<\/sup>]<\/p>\n<p><em>= 2a<sup>8<\/sup>\u00a0+ 12a<sup>6<\/sup>\u00a0\u2013 10a<sup>4<\/sup>\u00a0\u2013 4a<sup>2<\/sup>\u00a0+ 2<\/em><\/p>\n<p><strong>3. Find (a + b)<sup>\u00a04<\/sup>\u00a0\u2013 (a \u2013 b)<sup>\u00a04<\/sup>. Hence, evaluate (\u221a3 + \u221a2)<sup>4<\/sup>\u00a0\u2013 (\u221a3 \u2013 \u221a2)<sup>4<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Firstly, let us solve the given expression<\/p>\n<p>(a + b)<sup>\u00a04<\/sup>\u00a0\u2013 (a \u2013 b)<sup>\u00a04<\/sup><\/p>\n<p>The above expression can be expressed as,<\/p>\n<p>(a + b)<sup>\u00a04<\/sup>\u00a0\u2013 (a \u2013 b)<sup>\u00a04<\/sup>\u00a0= 2 [<sup>4<\/sup>C<sub>1<\/sub>\u00a0a<sup>3<\/sup>b<sup>1<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>3<\/sub>\u00a0a<sup>1<\/sup>b<sup>3<\/sup>]<\/p>\n<p>= 2 [4a<sup>3<\/sup>b + 4ab<sup>3<\/sup>]<\/p>\n<p>= 8 (a<sup>3<\/sup>b + ab<sup>3<\/sup>)<\/p>\n<p>Now,<\/p>\n<p>Let us evaluate the expression<\/p>\n<p>(\u221a3 + \u221a2)<sup>4<\/sup>\u00a0\u2013 (\u221a3 -\u221a2)<sup>4<\/sup><\/p>\n<p>So consider, a = \u221a3 and b = \u221a2 we get,<\/p>\n<p>(\u221a3 + \u221a2)<sup>4<\/sup>\u00a0\u2013 (\u221a3 -\u221a2)<sup>4<\/sup>\u00a0= 8 (a<sup>3<\/sup>b + ab<sup>3<\/sup>)<\/p>\n<p>= 8 [(\u221a3)<sup>3<\/sup>\u00a0(\u221a2) + (\u221a3) (\u221a2)<sup>3<\/sup>]<\/p>\n<p>= 8 [(3\u221a6) + (2\u221a6)]<\/p>\n<p>= 8 (5\u221a6)<\/p>\n<p><em>= 40\u221a6<\/em><\/p>\n<p><strong>4. Find (x + 1)<sup>\u00a06<\/sup>\u00a0+ (x \u2013 1)<sup>\u00a06<\/sup>. Hence, or otherwise, evaluate (\u221a2 + 1)<sup>6<\/sup>\u00a0+ (\u221a2 \u2013 1)<sup>6<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Firstly, let us solve the given expression<\/p>\n<p>(x + 1)<sup>\u00a06<\/sup>\u00a0+ (x \u2013 1)<sup>\u00a06<\/sup><\/p>\n<p>The above expression can be expressed as,<\/p>\n<p>(x + 1)<sup>\u00a06<\/sup>\u00a0+ (x \u2013 1)<sup>\u00a06<\/sup>\u00a0= 2 [<sup>6<\/sup>C<sub>0<\/sub>\u00a0x<sup>6<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>2<\/sub>\u00a0x<sup>4<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>4<\/sub>\u00a0x<sup>2<\/sup>\u00a0+\u00a0<sup>6<\/sup>C<sub>6<\/sub>\u00a0x<sup>0<\/sup>]<\/p>\n<p>= 2 [x<sup>6<\/sup>\u00a0+ 15x<sup>4<\/sup>\u00a0+ 15x<sup>2<\/sup>\u00a0+ 1]<\/p>\n<p>Now,<\/p>\n<p>Let us evaluate the expression<\/p>\n<p>(\u221a2 + 1)<sup>6<\/sup>\u00a0+ (\u221a2 \u2013 1)<sup>6<\/sup><\/p>\n<p>So consider x = \u221a2 then, we get,<\/p>\n<p>(\u221a2 + 1)<sup>6<\/sup>\u00a0+ (\u221a2 \u2013 1)<sup>6<\/sup>\u00a0= 2 [x<sup>6<\/sup>\u00a0+ 15x<sup>4<\/sup>\u00a0+ 15x<sup>2<\/sup>\u00a0+ 1]<\/p>\n<p>= 2 [(\u221a2)<sup>6<\/sup>\u00a0+ 15 (\u221a2)<sup>4<\/sup>\u00a0+ 15 (\u221a2)<sup>2<\/sup>\u00a0+ 1]<\/p>\n<p>= 2 [8 + 15 (4) + 15 (2) + 1]<\/p>\n<p>= 2 [8 + 60 + 30 + 1]<\/p>\n<p><em>= 198<\/em><\/p>\n<p><strong>5. Using the binomial theorem, evaluate each of the following:<\/strong><\/p>\n<p><strong>(i) (96)<sup>3<\/sup><\/strong><\/p>\n<p><strong>(ii) (102)<sup>5<\/sup><\/strong><\/p>\n<p><strong>(iii) (101)<sup>4<\/sup><\/strong><\/p>\n<p><strong>(iv) (98)<sup>5<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)\u00a0<\/strong>(96)<sup>3<\/sup><\/p>\n<p>We have,<\/p>\n<p>(96)<sup>3<\/sup><\/p>\n<p>Let us express the given expression as two different entities and apply the binomial theorem.<\/p>\n<p>(96)<sup>3<\/sup>\u00a0= (100 \u2013 4)<sup>3<\/sup><\/p>\n<p>=\u00a0<sup>3<\/sup>C<sub>0<\/sub>\u00a0(100)<sup>3<\/sup>\u00a0(4)<sup>0<\/sup>\u00a0\u2013\u00a0<sup>3<\/sup>C<sub>1<\/sub>\u00a0(100)<sup>2<\/sup>\u00a0(4)<sup>1<\/sup>\u00a0+\u00a0<sup>3<\/sup>C<sub>2<\/sub>\u00a0(100)<sup>1<\/sup>\u00a0(4)<sup>2<\/sup>\u00a0\u2013\u00a0<sup>3<\/sup>C<sub>3<\/sub>\u00a0(100)<sup>0<\/sup>\u00a0(4)<sup>3<\/sup><\/p>\n<p>= 1000000 \u2013 120000 + 4800 \u2013 64<\/p>\n<p><em>= 884736<\/em><\/p>\n<p><strong>(ii)\u00a0<\/strong>(102)<sup>5<\/sup><\/p>\n<p>We have,<\/p>\n<p>(102)<sup>5<\/sup><\/p>\n<p>Let us express the given expression as two different entities and apply the binomial theorem.<\/p>\n<p>(102)<sup>5<\/sup>\u00a0= (100 + 2)<sup>5<\/sup><\/p>\n<p>=\u00a0<sup>5<\/sup>C<sub>0<\/sub>\u00a0(100)<sup>5<\/sup>\u00a0(2)<sup>0<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>1<\/sub>\u00a0(100)<sup>4<\/sup>\u00a0(2)<sup>1<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(100)<sup>3<\/sup>\u00a0(2)<sup>2<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(100)<sup>2<\/sup>\u00a0(2)<sup>3<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(100)<sup>1<\/sup>\u00a0(2)<sup>4<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(100)<sup>0<\/sup>\u00a0(2)<sup>5<\/sup><\/p>\n<p>= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32<\/p>\n<p><em>= 11040808032<\/em><\/p>\n<p><strong>(iii)\u00a0<\/strong>(101)<sup>4<\/sup><\/p>\n<p>We have,<\/p>\n<p>(101)<sup>4<\/sup><\/p>\n<p>Let us express the given expression as two different entities and apply the binomial theorem.<\/p>\n<p>(101)<sup>4<\/sup>\u00a0= (100 + 1)<sup>4<\/sup><\/p>\n<p>=\u00a0<sup>4<\/sup>C<sub>0<\/sub>\u00a0(100)<sup>4<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>1<\/sub>\u00a0(100)<sup>3<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>2<\/sub>\u00a0(100)<sup>2<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>3<\/sub>\u00a0(100)<sup>1<\/sup>\u00a0+\u00a0<sup>4<\/sup>C<sub>4<\/sub>\u00a0(100)<sup>0<\/sup><\/p>\n<p>= 100000000 + 4000000 + 60000 + 400 + 1<\/p>\n<p><em>= 104060401<\/em><\/p>\n<p><strong>(iv)\u00a0<\/strong>(98)<sup>5<\/sup><\/p>\n<p>We have,<\/p>\n<p>(98)<sup>5<\/sup><\/p>\n<p>Let us express the given expression as two different entities and apply the binomial theorem.<\/p>\n<p>(98)<sup>5<\/sup>\u00a0= (100 \u2013 2)<sup>5<\/sup><\/p>\n<p>=\u00a0<sup>5<\/sup>C<sub>0<\/sub>\u00a0(100)<sup>5<\/sup>\u00a0(2)<sup>0<\/sup>\u00a0\u2013\u00a0<sup>5<\/sup>C<sub>1<\/sub>\u00a0(100)<sup>4<\/sup>\u00a0(2)<sup>1<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>2<\/sub>\u00a0(100)<sup>3<\/sup>\u00a0(2)<sup>2<\/sup>\u00a0\u2013\u00a0<sup>5<\/sup>C<sub>3<\/sub>\u00a0(100)<sup>2<\/sup>\u00a0(2)<sup>3<\/sup>\u00a0+\u00a0<sup>5<\/sup>C<sub>4<\/sub>\u00a0(100)<sup>1<\/sup>\u00a0(2)<sup>4<\/sup>\u00a0\u2013\u00a0<sup>5<\/sup>C<sub>5<\/sub>\u00a0(100)<sup>0<\/sup>\u00a0(2)<sup>5<\/sup><\/p>\n<p>= 10000000000 \u2013 1000000000 + 40000000 \u2013 800000 + 8000 \u2013 32<\/p>\n<p><em>= 9039207968<\/em><\/p>\n<p><strong>6. Using binomial theorem, prove that 2<sup>3n<\/sup>\u00a0\u2013 7n \u2013 1 is divisible by 49, where n \u2208 N.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>2<sup>3n<\/sup>\u00a0\u2013 7n \u2013 1<\/p>\n<p>So, 2<sup>3n<\/sup>\u00a0\u2013 7n \u2013 1 = 8<sup>n<\/sup>\u00a0\u2013 7n \u2013 1<\/p>\n<p>Now,<\/p>\n<p>8<sup>n<\/sup>\u00a0\u2013 7n \u2013 1<\/p>\n<p>8<sup>n<\/sup>\u00a0= 7n + 1<\/p>\n<p>= (1 + 7)<sup>\u00a0n<\/sup><\/p>\n<p>=\u00a0<sup>n<\/sup>C<sub>0<\/sub>\u00a0+\u00a0<sup>n<\/sup>C<sub>1<\/sub>\u00a0(7)<sup>1<\/sup>\u00a0+\u00a0<sup>n<\/sup>C<sub>2<\/sub>\u00a0(7)<sup>2<\/sup>\u00a0+\u00a0<sup>n<\/sup>C<sub>3<\/sub>\u00a0(7)<sup>3<\/sup>\u00a0+\u00a0<sup>n<\/sup>C<sub>4<\/sub>\u00a0(7)<sup>2<\/sup>\u00a0+\u00a0<sup>n<\/sup>C<sub>5<\/sub>\u00a0(7)<sup>1<\/sup>\u00a0+ \u2026 +\u00a0<sup>n<\/sup>C<sub>n<\/sub>\u00a0(7)<sup>\u00a0n<\/sup><\/p>\n<p>8<sup>n<\/sup>\u00a0= 1 + 7n + 49 [<sup>n<\/sup>C<sub>2<\/sub>\u00a0+\u00a0<sup>n<\/sup>C<sub>3<\/sub>\u00a0(7<sup>1<\/sup>) +\u00a0<sup>n<\/sup>C<sub>4\u00a0<\/sub>(7<sup>2<\/sup>) + \u2026 +\u00a0<sup>n<\/sup>C<sub>n<\/sub>\u00a0(7)<sup>\u00a0n-2<\/sup>]<\/p>\n<p>8<sup>n<\/sup>\u00a0\u2013 1 \u2013 7n = 49 (integer)<\/p>\n<p>So now,<\/p>\n<p>8<sup>n<\/sup>\u00a0\u2013 1 \u2013 7n is divisible by 49<\/p>\n<p>Or<\/p>\n<p><em>2<sup>3n<\/sup>\u00a0\u2013 1 \u2013 7n is divisible by 49.<\/em><\/p>\n<p>Hence proved.<\/p>\n<p>EXERCISE 18.2 PAGE NO: 18.37<\/p>\n<p><strong>1. Find the 11<sup>th<\/sup>\u00a0term from the beginning and the 11<sup>th<\/sup>\u00a0term from the end in the expansion of (2x \u2013 1\/x<sup>2<\/sup>)<sup>25<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(2x \u2013 1\/x<sup>2<\/sup>)<sup>25<\/sup><\/p>\n<p>The given expression contains 26 terms.<\/p>\n<p>So, the 11<sup>th<\/sup>\u00a0term from the end is the (26 \u2212 11 + 1)\u00a0<sup>th<\/sup>\u00a0term from the beginning.<\/p>\n<p>In other words, the 11<sup>th<\/sup>\u00a0term from the end is the 16<sup>th<\/sup>\u00a0term from the beginning.<\/p>\n<p>Then,<\/p>\n<p>T<sub>16<\/sub>\u00a0= T<sub>15+1<\/sub>\u00a0=\u00a0<sup>25<\/sup>C<sub>15<\/sub>\u00a0(2x)<sup>25-15<\/sup>\u00a0(-1\/x<sup>2<\/sup>)<sup>15<\/sup><\/p>\n<p>=\u00a0<sup>25<\/sup>C<sub>15<\/sub>\u00a0(2<sup>10<\/sup>) (x)<sup>10<\/sup>\u00a0(-1\/x<sup>30<\/sup>)<\/p>\n<p>= \u2013\u00a0<sup>25<\/sup>C<sub>15<\/sub>\u00a0(2<sup>10<\/sup>\u00a0\/ x<sup>20<\/sup>)<\/p>\n<p>Now, we shall find the 11<sup>th<\/sup>\u00a0term from the beginning.<\/p>\n<p>T<sub>11<\/sub>\u00a0= T<sub>10+1<\/sub>\u00a0=\u00a0<sup>25<\/sup>C<sub>10<\/sub>\u00a0(2x)<sup>25-10<\/sup>\u00a0(-1\/x<sup>2<\/sup>)<sup>10<\/sup><\/p>\n<p>=\u00a0<sup>25<\/sup>C<sub>10<\/sub>\u00a0(2<sup>15<\/sup>) (x)<sup>15<\/sup>\u00a0(1\/x<sup>20<\/sup>)<\/p>\n<p><em>=\u00a0<sup>25<\/sup>C<sub>10<\/sub>\u00a0(2<sup>15<\/sup>\u00a0\/ x<sup>5<\/sup>)<\/em><\/p>\n<p><strong>2. Find the 7<sup>th<\/sup>\u00a0term in the expansion of (3x<sup>2<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>)<sup>10<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(3x<sup>2<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>)<sup>10<\/sup><\/p>\n<p>Let us consider the 7<sup>th<\/sup>\u00a0term as T<sub>7<\/sub><\/p>\n<p>So,<\/p>\n<p>T<sub>7<\/sub>\u00a0= T<sub>6+1<\/sub><\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>6<\/sub>\u00a0(3x<sup>2<\/sup>)<sup>10-6<\/sup>\u00a0(-1\/x<sup>3<\/sup>)<sup>6<\/sup><\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>6<\/sub>\u00a0(3)<sup>4<\/sup>\u00a0(x)<sup>8<\/sup>\u00a0(1\/x<sup>18<\/sup>)<\/p>\n<p>= [10\u00d79\u00d78\u00d77\u00d781] \/ [4\u00d73\u00d72\u00d7x<sup>10<\/sup>]<\/p>\n<p>= 17010 \/ x<sup>10<\/sup><\/p>\n<p><em>\u2234 The 7<sup>th<\/sup>\u00a0term of the expression (3x<sup>2<\/sup>\u00a0\u2013 1\/x<sup>3<\/sup>)<sup>10<\/sup>\u00a0is 17010 \/ x<sup>10<\/sup>.<\/em><\/p>\n<p><strong>3.<\/strong>\u00a0<strong>Find the 5<sup>th<\/sup>\u00a0term in the expansion of (3x \u2013 1\/x<sup>2<\/sup>)<sup>10<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(3x \u2013 1\/x<sup>2<\/sup>)<sup>10<\/sup><\/p>\n<p>The 5<sup>th<\/sup>\u00a0term from the end is the (11 \u2013 5 + 1)th, is., 7<sup>th<\/sup>\u00a0term from the beginning.<\/p>\n<p>So,<\/p>\n<p>T<sub>7<\/sub>\u00a0= T<sub>6+1<\/sub><\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>6<\/sub>\u00a0(3x)<sup>10-6<\/sup>\u00a0(-1\/x<sup>2<\/sup>)<sup>6<\/sup><\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>6<\/sub>\u00a0(3)<sup>4<\/sup>\u00a0(x)<sup>4<\/sup>\u00a0(1\/x<sup>12<\/sup>)<\/p>\n<p>= [10\u00d79\u00d78\u00d77\u00d781] \/ [4\u00d73\u00d72\u00d7x<sup>8<\/sup>]<\/p>\n<p>= 17010 \/ x<sup>8<\/sup><\/p>\n<p><em>\u2234 The 5<sup>th<\/sup>\u00a0term of the expression (3x \u2013 1\/x<sup>2<\/sup>)<sup>10<\/sup>\u00a0is 17010 \/ x<sup>8<\/sup>.<\/em><\/p>\n<p><strong>4. Find the 8<sup>th<\/sup>\u00a0term in the expansion of (x<sup>3\/2<\/sup>\u00a0y<sup>1\/2<\/sup>\u00a0\u2013 x<sup>1\/2<\/sup>\u00a0y<sup>3\/2<\/sup>)<sup>10<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(x<sup>3\/2<\/sup>\u00a0y<sup>1\/2<\/sup>\u00a0\u2013 x<sup>1\/2<\/sup>\u00a0y<sup>3\/2<\/sup>)<sup>10<\/sup><\/p>\n<p>Let us consider the 8<sup>th<\/sup>\u00a0term as T<sub>8<\/sub><\/p>\n<p>So,<\/p>\n<p>T<sub>8<\/sub>\u00a0= T<sub>7+1<\/sub><\/p>\n<p>=\u00a0<sup>10<\/sup>C<sub>7<\/sub>\u00a0(x<sup>3\/2<\/sup>\u00a0y<sup>1\/2<\/sup>)<sup>10-7<\/sup>\u00a0(-x<sup>1\/2<\/sup>\u00a0y<sup>3\/2<\/sup>)<sup>7<\/sup><\/p>\n<p>= -[10\u00d79\u00d78]\/[3\u00d72] x<sup>9\/2<\/sup>\u00a0y<sup>3\/2<\/sup>\u00a0(x<sup>7\/2<\/sup>\u00a0y<sup>21\/2<\/sup>)<\/p>\n<p>= -120 x<sup>8<\/sup>y<sup>12<\/sup><\/p>\n<p><em>\u2234 The 8<sup>th<\/sup>\u00a0term of the expression (x<sup>3\/2<\/sup>\u00a0y<sup>1\/2<\/sup>\u00a0\u2013 x<sup>1\/2<\/sup>\u00a0y<sup>3\/2<\/sup>)<sup>10<\/sup>\u00a0is -120 x<sup>8<\/sup>y<sup>12<\/sup>.<\/em><\/p>\n<p><strong>5. Find the 7<sup>th<\/sup>\u00a0term in the expansion of (4x\/5 + 5\/2x)<sup>\u00a08<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(4x\/5 + 5\/2x)<sup>\u00a08<\/sup><\/p>\n<p>Let us consider the 7<sup>th<\/sup>\u00a0term as T<sub>7<\/sub><\/p>\n<p>So,<\/p>\n<p>T<sub>7<\/sub>\u00a0= T<sub>6+1<\/sub><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-21.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 21\" \/><\/p>\n<p><em>\u2234 The 7<sup>th<\/sup>\u00a0term of the expression (4x\/5 + 5\/2x)<sup>\u00a08<\/sup>\u00a0is 4375\/x<sup>4<\/sup>.<\/em><\/p>\n<p><strong>6. Find the 4<sup>th<\/sup>\u00a0term from the beginning and 4<sup>th<\/sup>\u00a0term from the end in the expansion of (x + 2\/x)<sup>\u00a09<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(x + 2\/x)<sup>\u00a09<\/sup><\/p>\n<p>Let\u00a0T<sub>r+1<\/sub>\u00a0be the 4th term from the end.<\/p>\n<p>Then,\u00a0T<sub>r+1<\/sub>\u00a0is (10 \u2212 4 + 1)th, i.e., 7th, the term from the beginning.<\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-22.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 22\" \/><\/p>\n<p><strong>7. Find the 4<sup>th<\/sup>\u00a0term from the end in the expansion of (4x\/5 \u2013 5\/2x)<sup>\u00a09<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(4x\/5 \u2013 5\/2x)<sup>\u00a09<\/sup><\/p>\n<p>Let T<sub>r+1<\/sub>\u00a0be the<sub>\u00a0<\/sub>4th term from the end of the given expression.<\/p>\n<p>Then, T<sub>r+1\u00a0<\/sub>is (10 \u2212 4 + 1)th term, i.e., the 7th term, from the beginning.<\/p>\n<p>T<sub>7<\/sub>\u00a0= T<sub>6+1<\/sub><\/p>\n<p><img title=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2020\/03\/rd-sharma-solutions-for-class-11-maths-chapter-18-binomial-theorem-image-23.png\" alt=\"RD Sharma Solutions for Class 11 Maths Chapter 18 \u2013 Binomial Theorem image - 23\" \/><\/p>\n<p><em>\u2234 The 4<sup>th<\/sup>\u00a0term from the end is 10500\/x<sup>3<\/sup>.<\/em><\/p>\n<h2><span class=\"ez-toc-section\" id=\"detailed-exercise-wise-explanation-with-listing-of-important-topics-in-the-exercise\"><\/span>Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-11-chapter-18-exercise-18a\"><\/span>RD Sharma Class 11 Chapter 18 Exercise 18A<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">In <\/span>Chapter 18A RD Sharma Class 11, <span style=\"font-weight: 400;\">the<\/span> <span style=\"font-weight: 400;\">students will be able to examine their performances by attempting <\/span><span style=\"font-weight: 400;\">objective-type questions that are associated with the binomial theorem for positive integral index and some important conclusions from the binomial theorem. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">After going through the solutions of RD Sharma Class 11 Chapter 18A the students will be able to understand the applications of the Binomial Theorem. The preparation style of the students will undergo a lot of changes which will be beneficial for their performances in the final Maths Exam. Hence, you must buy a copy for yourself right now.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"rd-sharma-class-11-chapter-18-exercise-18b\"><\/span>RD Sharma Class 11 Chapter 18 Exercise 18B<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><span style=\"font-weight: 400;\">In the <\/span>RD Sharma Class 11 Chapter 18 Exercise 18B Solutions <span style=\"font-weight: 400;\">you will be exposed to all kinds of tricky problems from the Binomial Theorem. Try to attempt all the questions from <\/span>RD Sharma Class 11 Chapter 18B <span style=\"font-weight: 400;\">without seeing the answers from Chapter 18 solutions. Analyze your performance and make a note of it.\u00a0<\/span><\/p>\n<h3><span class=\"ez-toc-section\" id=\"important-concepts-from-rd-sharma-solutions-class-11-maths-chapter-18\"><\/span><span style=\"font-weight: 400;\">Important concepts from RD Sharma Solutions Class 11 Maths Chapter 18<\/span><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Binomial theorem for positive integral index.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">Some important conclusions from the binomial theorem.<\/span><\/li>\n<li style=\"font-weight: 400;\"><span style=\"font-weight: 400;\">General terms and middle terms in a binomial expansion.<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"rd-sharma-solutions-class-11-maths-chapter-18\"><\/span>RD Sharma Solutions Class 11 Maths Chapter 18<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><span style=\"font-weight: 400;\">RD Sharma&#8217;s solution can play an important role in handling difficult problems from the Binomial Theorem. Collecting a copy of the RD Sharma Chapter 18 Solutions can help you to clear all the basic concepts of the Binomial Theorem. All the things will be easy to understand once you examine the relevant parts. <\/span><\/p>\n<p><span style=\"font-weight: 400;\">Some of the parts need special attention and the students will be able to keep track of it by referring to the solutions. There are different kinds of benefits for referring to <\/span>RD Sharma Chapter 18 Class 11\u00a0<span style=\"font-weight: 400;\">Solutions and now we will talk about those solutions. You must collect it after you finish your syllabus.\u00a0<\/span><\/p>\n<ul>\n<li><span style=\"font-weight: 400;\"> The students will be able to prepare themselves in a better way when they go through the contents of Maths Chapter 18.<\/span><span style=\"font-weight: 400;\"><br \/><\/span><\/li>\n<li><span style=\"font-weight: 400;\"> The students will be able to cross all the hurdles about the problems that are involved with Binomial Theorems.<\/span><\/li>\n<li>All the parts in connection with the basics of Binomial Theorems have been explained with proper presentations and if you examine the contents carefully you will be left with zero queries.\u00a0<\/li>\n<li><span style=\"font-weight: 400;\">You will get new ideas to improve your methods of preparation and that is the beauty of following <a href=\"http:\/\/cbse.nic.in\" target=\"_blank\" rel=\"noopener noreferrer\">CBSE<\/a> Chapter 18 Binomial Theorems.<\/span><\/li>\n<li><span style=\"font-weight: 400;\"> Shaping study materials becomes easier for you when you are ready with solutions.\u00a0<\/span><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"access-other-important-chapters-of-rd-sharma-solutions-class-11-maths\"><\/span>Access Other Important Chapters of RD Sharma Solutions Class 11 Maths\u00a0<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 1 &#8211; Sets<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 2 &#8211; Relations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 3 &#8211; Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 4 &#8211; Measurement of Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 5 &#8211; Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 6 &#8211; Graphs of Trigonometric Functions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 7 &#8211; Trigonometric Ratios of Compound Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 8 &#8211; Transformation Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 9 &#8211; Trigonometric Ratios of Multiple and Sub-Multiple Angles<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 10 &#8211; Sine and Cosine Formulae and Their Applications<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 11 &#8211; Trigonometric Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 12 &#8211; Mathematical Induction<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 13 &#8211; Complex Numbers<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 14 &#8211; Quadratic Equations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\">Chapter 15 &#8211; Linear Inequations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-16-permutations\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 16 &#8211; Permutations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-17-combinations\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 17 &#8211; Combinations<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-19-arithmetic-progressions\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 19 &#8211; Arithmetic Progressions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-20-geometric-progressions\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 20 &#8211; Geometric Progressions<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-21-some-special-series\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 21 &#8211; Some Special Series<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-22-brief-review-of-cartesian-system-of-rectangular-coordinates\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 22 &#8211; Brief Review of Cartesian System of Rectangular Coordinates<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-23-the-straight-lines\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 23 &#8211; The Straight Lines<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-24-the-circle\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 24 &#8211; The Circle<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-25-parabola\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 25 &#8211; Parabola<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-26-ellipse\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 26 &#8211; Ellipse<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-27-hyperbola\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 27 &#8211; Hyperbola<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-28-introduction-to-3d-coordinate-geometry\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 28 &#8211; Introduction To 3D Coordinate Geometry<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-29-limits\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 29 &#8211; Limits<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-30-derivatives\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 30 &#8211; Derivatives<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-31-mathematical-reasoning\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 31 &#8211; Mathematical Reasoning<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-32-statistics\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 32 &#8211; Statistics<\/a><\/li>\n<li><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-33-probability\/\" target=\"_blank\" rel=\"noopener noreferrer\">Chapter 33 &#8211; Probability<\/a><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-class-11-maths-chapter-18\"><\/span>FAQs on RD Sharma Solutions Class 11 Maths Chapter 18<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1629897420623\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-18\"><\/span>From where can I download the PDF of\u00a0RD Sharma Solutions Class 11 Maths Chapter 18?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link in the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629897439045\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-18\"><\/span>How much does it cost to download the PDF of\u00a0RD Sharma Solutions Class 11 Maths Chapter 18?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.\u00a0<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1629897472394\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-solutions-class-11-maths-chapter-18-pdf-offline\"><\/span>Can I access the\u00a0RD Sharma Solutions Class 11 Maths Chapter 18 PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online you can access it offline as well.\u00a0<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem:\u00a0It is RD Sharma Solutions Class 11 Maths Chapter 18 Binomial Theorem in which all the parts of the blog are associated with the fundamental aspects of Binomial Theorem like the benefits of following RD Sharma Solutions Class 11 Maths Chapter 18 Maths Solutions along &#8230; <a title=\"RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem (Updated For 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-18\/\" aria-label=\"More on RD Sharma Solutions Class 11 Maths Chapter 18 &#8211; Binomial Theorem (Updated For 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":64338,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62998"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=62998"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62998\/revisions"}],"predecessor-version":[{"id":518928,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/62998\/revisions\/518928"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/64338"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=62998"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=62998"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=62998"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}