{"id":67944,"date":"2023-09-16T14:22:00","date_gmt":"2023-09-16T08:52:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67944"},"modified":"2023-12-01T10:17:23","modified_gmt":"2023-12-01T04:47:23","slug":"rd-sharma-solutions-class-11-maths-chapter-12-exercise-12-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-12-exercise-12-2\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 12 Exercise 12.2 (Updated for 2023-24)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122651\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-12-Exercise-12.2.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-12-Exercise-12.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/RD-Sharma-Class-11-Solutions-Chapter-12-Exercise-12.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2:<\/strong> Acing your Class 11 Maths exams is made easier, all thanks to the <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> You can clear your doubts and strengthen you basics with this ultimate guide. Download the Free PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-12-mathematical-induction\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 12<\/a> Exercise 12.2 today.<\/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-69d74fa2f21a8\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-12-exercise-12-2\/#from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-12-ex-122\" title=\"From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 12 Ex 12.2?\">From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 12 Ex 12.2?<\/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-12-exercise-12-2\/#how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-12-exercise-122\" title=\"How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 12 Exercise 12.2?\">How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 12 Exercise 12.2?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-solutions-class-11-maths-chapter-12-exercise-122-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2 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\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-12-Ex-12.2-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2<\/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\/01\/RD-Sharma-Solutions-Class-11-Maths-Chapter-12-Ex-12.2-1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-class-11-maths-chapter-12-exercise-122\"><\/span>Access answers of RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Prove the following by the principle of mathematical induction.<\/strong><\/p>\n<p><strong>1. 1 + 2 + 3 + \u2026 + n = n (n +1)\/2, i.e., the sum of the first n natural numbers is n (n + 1)\/2.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider P (n) = 1 + 2 + 3 + \u2026.. + n = n (n +1)\/2<\/p>\n<p>For, n = 1<\/p>\n<p>LHS of P (n) = 1<\/p>\n<p>RHS of P (n) =\u00a01 (1+1)\/2 = 1<\/p>\n<p>So, LHS = RHS<\/p>\n<p>Since, P (n) is true for n = 1<\/p>\n<p>Let us consider P (n) to be true for n = k, so<\/p>\n<p>1 + 2 + 3 + \u2026. + k = k (k+1)\/2 \u2026 (i)<\/p>\n<p>Now,<\/p>\n<p>(1 + 2 + 3 + \u2026 + k) + (k + 1) = k (k+1)\/2 + (k+1)<\/p>\n<p>= (k + 1) (k\/2 + 1)<\/p>\n<p>= [(k + 1) (k + 2)] \/ 2<\/p>\n<p>= [(k+1) [(k+1) + 1]] \/ 2<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>P (n) is true for all n \u2208 N<\/p>\n<p>So, by the principle of Mathematical Induction,<\/p>\n<p>P (n) = 1 + 2 + 3 + \u2026.. + n = n (n +1)\/2 is true for all n \u2208 N.<\/p>\n<p><strong>2. 1<sup>2<\/sup>\u00a0+ 2<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ \u2026 + n<sup>2<\/sup>\u00a0= [n (n+1) (2n+1)]\/6<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let us consider P (n) = 1<sup>2<\/sup>\u00a0+ 2<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ \u2026 + n<sup>2<\/sup>\u00a0= [n (n+1) (2n+1)]\/6<\/p>\n<p>For, n = 1<\/p>\n<p>P (1) = [1 (1+1) (2+1)]\/6<\/p>\n<p>1 = 1<\/p>\n<p>P (n) is true for n = 1<\/p>\n<p>Let P (n) is true for n = k, so<\/p>\n<p>P (k): 1<sup>2<\/sup>\u00a0+ 2<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ \u2026 + k<sup>2<\/sup>\u00a0= [k (k+1) (2k+1)]\/6<\/p>\n<p>Let\u2019s check for P (n) = k + 1, so<\/p>\n<p>P (k) = 1<sup>2<\/sup>\u00a0+ 2<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ \u2013 \u2013 \u2013 \u2013 \u2013 + k<sup>2<\/sup>\u00a0+ (k + 1)<sup>2<\/sup>\u00a0= [k + 1 (k+2) (2k+3)] \/6<\/p>\n<p>= 1<sup>2<\/sup>\u00a0+ 2<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ \u2013 \u2013 \u2013 \u2013 \u2013 + k<sup>2<\/sup>\u00a0+ (k + 1)<sup>2<\/sup><\/p>\n<p>= [k + 1 (k+2) (2k+3)] \/6 + (k + 1)<sup>2<\/sup><\/p>\n<p>= (k +1) [(2k<sup>2<\/sup>\u00a0+ k)\/6 + (k + 1)\/1]<\/p>\n<p>= (k +1) [2k<sup>2<\/sup>\u00a0+ k + 6k + 6]\/6<\/p>\n<p>= (k +1) [2k<sup>2<\/sup>\u00a0+ 7k + 6]\/6<\/p>\n<p>= (k +1) [2k<sup>2<\/sup>\u00a0+ 4k + 3k + 6]\/6<\/p>\n<p>= (k +1) [2k(k + 2) + 3(k + 2)]\/6<\/p>\n<p>= [(k +1) (2k + 3) (k + 2)] \/ 6<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>3. 1 + 3 + 3<sup>2<\/sup>\u00a0+ \u2026 + 3<sup>n-1<\/sup>\u00a0= (3<sup>n<\/sup>\u00a0\u2013 1)\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 1 + 3 + 3<sup>2<\/sup>\u00a0+ \u2013 \u2013 \u2013 \u2013 + 3<sup>n \u2013 1<\/sup>\u00a0= (3<sup>n<\/sup>\u00a0\u2013 1)\/2<\/p>\n<p>Now, for n = 1<\/p>\n<p>P (1) = 1 = (3<sup>1<\/sup>\u00a0\u2013 1)\/2 = 2\/2\u00a0=1<\/p>\n<p>P (n) is true for n = 1<\/p>\n<p>Now, let\u2019s check for P (n) is true for n = k<\/p>\n<p>P (k) = 1 + 3 + 3<sup>2<\/sup>\u00a0+ \u2013 \u2013 \u2013 \u2013 + 3<sup>k \u2013 1<\/sup>\u00a0= (3<sup>k<\/sup>\u00a0\u2013 1)\/2 \u2026 (i)<\/p>\n<p>Now, we have to show that P (n) is true for n = k + 1<\/p>\n<p>P (k + 1) = 1 + 3 + 3<sup>2<\/sup>\u00a0+ \u2013 \u2013 \u2013 \u2013 + 3<sup>k<\/sup>\u00a0= (3<sup>k+1<\/sup>\u00a0\u2013 1)\/2<\/p>\n<p>Then, {1 + 3 + 3<sup>2<\/sup>\u00a0+ \u2013 \u2013 \u2013 \u2013 + 3<sup>k \u2013 1<\/sup>} + 3<sup>k + 1 \u2013 1<\/sup><\/p>\n<p>= (3k \u2013 1)\/2 + 3<sup>k<\/sup>\u00a0using equation (i)<\/p>\n<p>= (3k \u2013 1 + 2\u00d73<sup>k<\/sup>)\/2<\/p>\n<p>= (3\u00d73 k \u2013 1)\/2<\/p>\n<p>= (3<sup>k+1<\/sup>\u00a0\u2013 1)\/2<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>4. 1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/n(n+1) = n\/(n+1)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/n(n+1) = n\/(n+1)<\/p>\n<p>For, n = 1<\/p>\n<p>P (n) = 1\/1.2 = 1\/1+1<\/p>\n<p>1\/2 = 1\/2<\/p>\n<p>P (n) is true for n = 1<\/p>\n<p>Let\u2019s check for P (n) is true for n = k,<\/p>\n<p>1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/k(k+1) + k\/(k+1) (k+2) = (k+1)\/(k+2)<\/p>\n<p>Then,<\/p>\n<p>1\/1.2 + 1\/2.3 + 1\/3.4 + \u2026 + 1\/k(k+1) + k\/(k+1) (k+2)<\/p>\n<p>= 1\/(k+1)\/(k+2) + k\/(k+1)<\/p>\n<p>= 1\/(k+1) [k(k+2)+1]\/(k+2)<\/p>\n<p>= 1\/(k+1) [k<sup>2<\/sup>\u00a0+ 2k + 1]\/(k+2)<\/p>\n<p>=1\/(k+1) [(k+1) (k+1)]\/(k+2)<\/p>\n<p>= (k+1) \/ (k+2)<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>5. 1 + 3 + 5 + \u2026 + (2n \u2013 1) = n<sup>2<\/sup>, i.e., the sum of the first n odd natural numbers is n<sup>2<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): 1 + 3 + 5 + \u2026 + (2n \u2013 1) = n<sup>2<\/sup><\/p>\n<p>Let us check if P (n) is true for n = 1<\/p>\n<p>P (1) = 1 =1<sup>2<\/sup><\/p>\n<p>1 = 1<\/p>\n<p>P (n) is true for n = 1<\/p>\n<p>Now, Let\u2019s check that P (n) is true for n = k<\/p>\n<p>P (k) = 1 + 3 + 5 + \u2026 + (2k \u2013 1) = k<sup>2<\/sup>\u00a0\u2026 (i)<\/p>\n<p>We have to show that<\/p>\n<p>1 + 3 + 5 + \u2026 + (2k \u2013 1) + 2(k + 1) \u2013 1 = (k + 1)<sup>2<\/sup><\/p>\n<p>Now,<\/p>\n<p>1 + 3 + 5 + \u2026 + (2k \u2013 1) + 2(k + 1) \u2013 1<\/p>\n<p>= k<sup>2<\/sup>\u00a0+ (2k + 1)<\/p>\n<p>= k<sup>2<\/sup>\u00a0+ 2k + 1<\/p>\n<p>= (k + 1)<sup>2<\/sup><\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>6. 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3n-1) (3n+2) = n\/(6n+4)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3n-1) (3n+2) = n\/(6n+4)<\/p>\n<p>Let us check if P (n) is true for n = 1<\/p>\n<p>P (1): 1\/2.5 = 1\/6.1+4 =&gt; 1\/10 = 1\/10<\/p>\n<p>P (1) is true.<\/p>\n<p>Now,<\/p>\n<p>Let us check that P (k) is true and prove that P (k + 1) is true.<\/p>\n<p>P (k): 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3k-1) (3k+2) = k\/(6k+4)<\/p>\n<p>P (k +1): 1\/2.5 + 1\/5.8 + 1\/8.11 + \u2026 + 1\/(3k-1)(3k+2) + 1\/(3k+3-1)(3k+3+2)<\/p>\n<p>: k\/(6k+4) + 1\/(3k+2)(3k+5)<\/p>\n<p>: [k(3k+5)+2] \/ [2(3k+2)(3k+5)]<\/p>\n<p>: (k+1) \/ (6(k+1)+4)<\/p>\n<p>P (k + 1) is true.<\/p>\n<p>Hence proved by mathematical induction.<\/p>\n<p><strong>7. 1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3n-2)(3n+1) = n\/3n+1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3n-2)(3n+1) = n\/3n+1<\/p>\n<p>Let us check for n = 1<\/p>\n<p>P (1): 1\/1.4 = 1\/4<\/p>\n<p>1\/4 = 1\/4<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check for P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k) = 1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3k-2)(3k+1) = k\/3k+1 \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>[1\/1.4 + 1\/4.7 + 1\/7.10 + \u2026 + 1\/(3k-2)(3k+1)]+ 1\/(3k+1)(3k+4)<\/p>\n<p>= k\/(3k+1) + 1\/(3k+1)(3k+4)<\/p>\n<p>= 1\/(3k+1) [k\/1 + 1\/(3k+4)]<\/p>\n<p>= 1\/(3k+1) [k(3k+4)+1]\/(3k+4)<\/p>\n<p>= 1\/(3k+1) [3k<sup>2<\/sup>\u00a0+ 4k + 1]\/ (3k+4)<\/p>\n<p>= 1\/(3k+1) [3k<sup>2<\/sup>\u00a0+ 3k+k+1]\/(3k+4)<\/p>\n<p>= [3k(k+1) + (k+1)] \/ [(3k+4) (3k+1)]<\/p>\n<p>= [(3k+1)(k+1)] \/ [(3k+4) (3k+1)]<\/p>\n<p>= (k+1) \/ (3k+4)<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>8. 1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2n+1)(2n+3) = n\/3(2n+3)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2n+1)(2n+3) = n\/3(2n+3)<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 1\/3.5 = 1\/3(2.1+3)<\/p>\n<p>: 1\/15 = 1\/15<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k) = 1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2k+1)(2k+3) = k\/3(2k+3) \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2k+1)(2k+3) + 1\/[2(k+1)+1][2(k+1)+3]<\/p>\n<p>1\/3.5 + 1\/5.7 + 1\/7.9 + \u2026 + 1\/(2k+1)(2k+3) + 1\/(2k+3)(2k+5)<\/p>\n<p>Now substituting the value of P (k), we get<\/p>\n<p>= k\/3(2k+3) + 1\/(2k+3)(2k+5)<\/p>\n<p>= [k(2k+5)+3] \/ [3(2k+3)(2k+5)]<\/p>\n<p>= (k+1) \/ [3(2(k+1)+3)]<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>9. 1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4n-1)(4n+3) = n\/3(4n+3)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4n-1)(4n+3) = n\/3(4n+3)<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 1\/3.7 = 1\/(4.1-1)(4+3)<\/p>\n<p>: 1\/21 = 1\/21<\/p>\n<p>P (n) is true for n =1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4k-1)(4k+3) = k\/3(4k+3) \u2026. (i)<\/p>\n<p>So,<\/p>\n<p>1\/3.7 + 1\/7.11 + 1\/11.15 + \u2026 + 1\/(4k-1)(4k+3) + 1\/(4k+3)(4k+7)<\/p>\n<p>Substituting the value of P (k), we get<\/p>\n<p>= k\/(4k+3) + 1\/(4k+3)(4k+7)<\/p>\n<p>= 1\/(4k+3) [k(4k+7)+3] \/ [3(4k+7)]<\/p>\n<p>= 1\/(4k+3) [4k<sup>2<\/sup>\u00a0+ 7k +3]\/ [3(4k+7)]<\/p>\n<p>= 1\/(4k+3) [4k<sup>2<\/sup>\u00a0+ 3k+4k+3] \/ [3(4k+7)]<\/p>\n<p>= 1\/(4k+3) [4k(k+1)+3(k+1)]\/ [3(4k+7)]<\/p>\n<p>= 1\/(4k+3) [(4k+3)(k+1)] \/ [3(4k+7)]<\/p>\n<p>= (k+1) \/ [3(4k+7)]<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208\u00a0N.<\/p>\n<p><strong>10. 1.2 + 2.2<sup>2<\/sup>\u00a0+ 3.2<sup>3<\/sup>\u00a0+ \u2026 + n.2<sup>n\u00a0<\/sup>= (n\u20131) 2<sup>n + 1<\/sup>\u00a0+ 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) =\u00a01.2 + 2.2<sup>2<\/sup>\u00a0+ 3.2<sup>3<\/sup>\u00a0+ \u2026 + n.2<sup>n\u00a0<\/sup>= (n\u20131) 2<sup>n + 1<\/sup>\u00a0+ 2<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1):1.2 = 0.2<sup>0<\/sup>\u00a0+ 2<\/p>\n<p>: 2 = 2<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k):\u00a01.2 + 2.2<sup>2<\/sup>\u00a0+ 3.2<sup>3<\/sup>\u00a0+ \u2026 + k.2<sup>k\u00a0<\/sup>= (k\u20131) 2<sup>k + 1<\/sup>\u00a0+ 2 \u2026. (i)<\/p>\n<p>So,<\/p>\n<p>{1.2 + 2.2<sup>2<\/sup>\u00a0+ 3.2<sup>3<\/sup>\u00a0+ \u2026 + k.2<sup>k<\/sup>} + (k + 1)2<sup>k + 1<\/sup><\/p>\n<p>Now, substituting the value of P (k), we get<\/p>\n<p>= [(k \u2013 1)2<sup>k + 1<\/sup>\u00a0+ 2] + (k + 1)2<sup>k + 1<\/sup>\u00a0using equation (i)<\/p>\n<p>= (k \u2013 1)2<sup>k + 1<\/sup>\u00a0+ 2 + (k + 1)2<sup>k + 1<\/sup><\/p>\n<p>= 2<sup>k + 1<\/sup>(k \u2013 1 + k + 1) + 2<\/p>\n<p>= 2<sup>k + 1<\/sup>\u00a0\u00d7 2k + 2<\/p>\n<p>= k \u00d7 2<sup>k + 2<\/sup>\u00a0+ 2<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>11. 2 + 5 + 8 + 11 + \u2026 + (3n \u2013 1) = 1\/2 n (3n + 1)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n) = 2 + 5 + 8 + 11 + \u2026 + (3n \u2013 1) = 1\/2 n (3n + 1)<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 2 = 1\/2 \u00d7 1 \u00d7 4<\/p>\n<p>: 2 = 2<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k) = 2 + 5 + 8 + 11 + \u2026 + (3k \u2013 1) = 1\/2 k (3k + 1) \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>2 + 5 + 8 + 11 + \u2026 + (3k \u2013 1) + (3k + 2)<\/p>\n<p>Now, substituting the value of P (k), we get,<\/p>\n<p>= 1\/2 \u00d7 k (3k + 1) + (3k + 2) by using equation (i)<\/p>\n<p>= [3k<sup>2<\/sup>\u00a0+ k + 2 (3k + 2)] \/ 2<\/p>\n<p>= [3k<sup>2<\/sup>\u00a0+ k + 6k + 2] \/ 2<\/p>\n<p>= [3k<sup>2<\/sup>\u00a0+ 7k + 2] \/ 2<\/p>\n<p>= [3k<sup>2<\/sup>\u00a0+ 4k + 3k + 2] \/ 2<\/p>\n<p>= [3k (k+1) + 4(k+1)] \/ 2<\/p>\n<p>= [(k+1) (3k+4)] \/2<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>12. 1.3 + 2.4 + 3.5 + \u2026 + n. (n+2) = 1\/6 n (n+1) (2n+7)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): 1.3 + 2.4 + 3.5 + \u2026 + n. (n+2) = 1\/6 n (n+1) (2n+7)<\/p>\n<p>Let us check for n = 1<\/p>\n<p>P (1): 1.3 = 1\/6 \u00d7 1 \u00d7 2 \u00d7 9<\/p>\n<p>: 3 = 3<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 1.3 + 2.4 + 3.5 + \u2026 + k. (k+2) = 1\/6 k (k+1) (2k+7) \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>1.3 + 2.4 + 3.5 + \u2026 + k. (k+2) + (k+1) (k+3)<\/p>\n<p>Now, substituting the value of P (k), we get<\/p>\n<p>= 1\/6 k (k+1) (2k+7) + (k+1) (k+3) by using equation (i)<\/p>\n<p>= (k+1) [{k(2k+7)\/6} + {(k+3)\/1}]<\/p>\n<p>= (k+1) [(2k<sup>2<\/sup>\u00a0+ 7k + 6k + 18)] \/ 6<\/p>\n<p>= (k+1) [2k<sup>2<\/sup>\u00a0+ 13k + 18] \/ 6<\/p>\n<p>= (k+1) [2k<sup>2<\/sup>\u00a0+ 9k + 4k + 18] \/ 6<\/p>\n<p>= (k+1) [2k(k+2) + 9(k+2)] \/ 6<\/p>\n<p>= (k+1) [(2k+9) (k+2)] \/ 6<\/p>\n<p>= 1\/6 (k+1) (k+2) (2k+9)<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>13. 1.3 + 3.5 + 5.7 + \u2026 + (2n \u2013 1) (2n + 1) = n(4n<sup>2<\/sup>\u00a0+ 6n \u2013 1)\/3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a01.3 + 3.5 + 5.7 + \u2026 + (2n \u2013 1) (2n + 1) = n(4n<sup>2<\/sup>\u00a0+ 6n \u2013 1)\/3<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): (2.1 \u2013 1) (2.1 + 1) = 1(4.1<sup>2<\/sup>\u00a0+ 6.1 -1)\/3<\/p>\n<p>: 1\u00d73 = 1(4+6-1)\/3<\/p>\n<p>: 3 = 3<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 1.3 + 3.5 + 5.7 + \u2026 + (2k \u2013 1) (2k + 1) = k(4k<sup>2<\/sup>\u00a0+ 6k \u2013 1)\/3 \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>1.3 + 3.5 + 5.7 + \u2026 + (2k \u2013 1) (2k + 1) + (2k + 1) (2k + 3)<\/p>\n<p>Now, substituting the value of P (k), we get<\/p>\n<p>= k(4k<sup>2<\/sup>\u00a0+ 6k \u2013 1)\/3 + (2k + 1) (2k + 3) by using equation (i)<\/p>\n<p>= [k(4k<sup>2<\/sup>\u00a0+ 6k-1) + 3 (4k<sup>2<\/sup>\u00a0+ 6k + 2k + 3)] \/ 3<\/p>\n<p>= [4k<sup>3<\/sup>\u00a0+ 6k<sup>2<\/sup>\u00a0\u2013 k + 12k<sup>2<\/sup>\u00a0+ 18k + 6k + 9] \/3<\/p>\n<p>= [4k<sup>3<\/sup>\u00a0+ 18k<sup>2<\/sup>\u00a0+ 23k + 9] \/3<\/p>\n<p>= [4k<sup>3<\/sup>\u00a0+ 4k<sup>2<\/sup>\u00a0+ 14k<sup>2<\/sup>\u00a0+ 14k +9k + 9] \/3<\/p>\n<p>= [(k+1) (4k<sup>2<\/sup>\u00a0+ 8k +4 + 6k + 6 \u2013 1)] \/ 3<\/p>\n<p>= [(k+1) 4[(k+1)<sup>2<\/sup>\u00a0+ 6(k+1) -1]] \/3<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>14. 1.2 + 2.3 + 3.4 + \u2026 + n(n+1) = [n (n+1) (n+2)] \/ 3<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a01.2 + 2.3 + 3.4 + \u2026 + n(n+1) = [n (n+1) (n+2)] \/ 3<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 1(1+1) = [1(1+1) (1+2)] \/3<\/p>\n<p>: 2 = 2<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 1.2 + 2.3 + 3.4 + \u2026 + k(k+1) = [k (k+1) (k+2)] \/ 3 \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>1.2 + 2.3 + 3.4 + \u2026 + k(k+1) + (k+1) (k+2)<\/p>\n<p>Now, substituting the value of P (k), we get<\/p>\n<p>= [k (k+1) (k+2)] \/ 3 + (k+1) (k+2) by using equation (i)<\/p>\n<p>= (k+2) (k+1) [k\/2 + 1]<\/p>\n<p>= [(k+1) (k+2) (k+3)] \/3<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>15. 1\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2<sup>n<\/sup>\u00a0= 1 \u2013 1\/2<sup>n<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a01\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2<sup>n<\/sup>\u00a0= 1 \u2013 1\/2<sup>n<\/sup><\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 1\/2<sup>1<\/sup>\u00a0= 1 \u2013 1\/2<sup>1<\/sup><\/p>\n<p>: 1\/2 = 1\/2<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>Let P (k):\u00a01\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2<sup>k<\/sup>\u00a0= 1 \u2013 1\/2<sup>k<\/sup>\u00a0\u2026 (i)<\/p>\n<p>So,<\/p>\n<p>1\/2 + 1\/4 + 1\/8 + \u2026 + 1\/2<sup>k<\/sup>\u00a0+ 1\/2<sup>k+1<\/sup><\/p>\n<p>Now, substituting the value of P (k), we get<\/p>\n<p>= 1 \u2013 1\/2<sup>k\u00a0<\/sup>+ 1\/2<sup>k+1<\/sup>\u00a0by using equation (i)<\/p>\n<p>= 1 \u2013 ((2-1)\/2<sup>k+1<\/sup>)<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>16. 1<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>\u00a0+ \u2026 + (2n \u2013 1)<sup>2<\/sup>\u00a0= 1\/3 n (4n<sup>2<\/sup>\u00a0\u2013 1)<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a01<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>\u00a0+ \u2026 + (2n \u2013 1)<sup>2<\/sup>\u00a0= 1\/3 n (4n<sup>2<\/sup>\u00a0\u2013 1)<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): (2.1 \u2013 1)<sup>2<\/sup>\u00a0= 1\/3 \u00d7 1 \u00d7 (4 \u2013 1)<\/p>\n<p>: 1 = 1<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 1<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>\u00a0+ \u2026 + (2k \u2013 1)<sup>2<\/sup>\u00a0= 1\/3 k (4k<sup>2<\/sup>\u00a0\u2013 1) \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>1<sup>2<\/sup>\u00a0+ 3<sup>2<\/sup>\u00a0+ 5<sup>2<\/sup>\u00a0+ \u2026 + (2k \u2013 1)<sup>2<\/sup>\u00a0+ (2k + 1)<sup>2<\/sup><\/p>\n<p>Now, substituting the value of P (k), we get,<\/p>\n<p>= 1\/3 k (4k<sup>2<\/sup>\u00a0\u2013 1) + (2k + 1)<sup>2<\/sup>\u00a0by using equation (i)<\/p>\n<p>= 1\/3 k (2k + 1) (2k \u2013 1) + (2k + 1)<sup>2<\/sup><\/p>\n<p>= (2k + 1) [{k(2k-1)\/3} + (2k+1)]<\/p>\n<p>= (2k + 1) [2k<sup>2<\/sup>\u00a0\u2013 k + 3(2k+1)] \/ 3<\/p>\n<p>= (2k + 1) [2k<sup>2<\/sup>\u00a0\u2013 k + 6k + 3] \/ 3<\/p>\n<p>= [(2k+1) 2k<sup>2<\/sup>\u00a0+ 5k + 3] \/3<\/p>\n<p>= [(2k+1) (2k(k+1)) + 3 (k+1)] \/3<\/p>\n<p>= [(2k+1) (2k+3) (k+1)] \/3<\/p>\n<p>= (k+1)\/3 [4k<sup>2<\/sup>\u00a0+ 6k + 2k + 3]<\/p>\n<p>= (k+1)\/3 [4k<sup>2<\/sup>\u00a0+ 8k \u2013 1]<\/p>\n<p>= (k+1)\/3 [4(k+1)<sup>2<\/sup>\u00a0\u2013 1]<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>17. a + ar + ar<sup>2<\/sup>\u00a0+ \u2026 + ar<sup>n \u2013 1<\/sup>\u00a0= a [(r<sup>n<\/sup>\u00a0\u2013 1)\/(r \u2013 1)], r \u2260 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a0a + ar + ar<sup>2<\/sup>\u00a0+ \u2026 + ar<sup>n \u2013 1<\/sup>\u00a0= a [(r<sup>n<\/sup>\u00a0\u2013 1)\/(r \u2013 1)]<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): a = a (r<sup>1<\/sup>\u00a0\u2013 1)\/(r-1)<\/p>\n<p>: a = a<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): a + ar + ar<sup>2<\/sup>\u00a0+ \u2026 + ar<sup>k \u2013 1<\/sup>\u00a0= a [(r<sup>k<\/sup>\u00a0\u2013 1)\/(r \u2013 1)] \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>a + ar + ar<sup>2<\/sup>\u00a0+ \u2026 + ar<sup>k \u2013 1<\/sup>\u00a0+ ar<sup>k<\/sup><\/p>\n<p>Now, substituting the value of P (k), we get,<\/p>\n<p>= a [(r<sup>k<\/sup>\u00a0\u2013 1)\/(r \u2013 1)] + ar<sup>k<\/sup>\u00a0by using equation (i)<\/p>\n<p>= a[r<sup>k<\/sup>\u00a0\u2013 1 + r<sup>k<\/sup>(r-1)] \/ (r-1)<\/p>\n<p>= a[r<sup>k<\/sup>\u00a0\u2013 1 + r<sup>k+1<\/sup>\u00a0\u2013 r<sup>\u2011k<\/sup>] \/ (r-1)<\/p>\n<p>= a[r<sup>k+1<\/sup>\u00a0\u2013 1] \/ (r-1)<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>18. a + (a + d) + (a + 2d) + \u2026 + (a + (n-1)d) = n\/2 [2a + (n-1)d]<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a0a + (a + d) + (a + 2d) + \u2026 + (a + (n-1)d) = n\/2 [2a + (n-1)d]<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): a = \u00bd [2a + (1-1)d]<\/p>\n<p>: a = a<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): a + (a + d) + (a + 2d) + \u2026 + (a + (k-1)d) = k\/2 [2a + (k-1)d] \u2026 (i)<\/p>\n<p>So,<\/p>\n<p>a + (a + d) + (a + 2d) + \u2026 + (a + (k-1)d) + (a + (k)d)<\/p>\n<p>Now, substituting the value of P (k), we get,<\/p>\n<p>= k\/2 [2a + (k-1)d] + (a + kd) by using equation (i)<\/p>\n<p>= [2ka + k(k-1)d + 2(a+kd)] \/ 2<\/p>\n<p>= [2ka + k<sup>2<\/sup>d \u2013 kd + 2a + 2kd] \/ 2<\/p>\n<p>= [2ka + 2a + k<sup>2<\/sup>d + kd] \/ 2<\/p>\n<p>= [2a(k+1) + d(k<sup>2<\/sup>\u00a0+ k)] \/ 2<\/p>\n<p>= (k+1)\/2 [2a + kd]<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>19. 5<sup>2n<\/sup>\u00a0\u2013 1 is divisible by 24 for all n\u00a0\u03f5\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n):\u00a05<sup>2n<\/sup>\u00a0\u2013 1 is divisible by 24<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 5<sup>2<\/sup>\u00a0\u2013 1 = 25 \u2013 1 = 24<\/p>\n<p>P (n) is true for n = 1. Where P (n) is divisible by 24<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 5<sup>2k<\/sup>\u00a0\u2013 1 is divisible by 24<\/p>\n<p>: 5<sup>2k<\/sup>\u00a0\u2013 1 = 24\u03bb \u2026 (i)<\/p>\n<p>We have to prove,<\/p>\n<p>5<sup>2k + 1<\/sup>\u00a0\u2013 1\u00a0is divisible by 24<\/p>\n<p>5<sup>2(k + 1)<\/sup>\u00a0\u2013 1 = 24\u03bc<\/p>\n<p>So,<\/p>\n<p>= 5<sup>2(k + 1)<\/sup>\u00a0\u2013 1<\/p>\n<p>= 5<sup>2k<\/sup>.5<sup>2<\/sup>\u00a0\u2013\u00a01<\/p>\n<p>= 25.5<sup>2k<\/sup>\u00a0\u2013 1<\/p>\n<p>= 25.(24\u03bb\u00a0+ 1) \u2013 1 by using equation (1)<\/p>\n<p>= 25.24\u03bb + 24<\/p>\n<p>= 24\u03bb<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>20. 3<sup>2n<\/sup>\u00a0+ 7 is divisible by 8 for all n\u00a0\u03f5\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): 3<sup>2n<\/sup>\u00a0+ 7 is divisible by 8<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 3<sup>2<\/sup>\u00a0+ 7 = 9 + 7 = 16<\/p>\n<p>P (n) is true for n = 1. Where P (n) is divisible by 8<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 3<sup>2k<\/sup>\u00a0+ 7 is divisible by 8<\/p>\n<p>: 3<sup>2k<\/sup>\u00a0+ 7 = 8\u03bb<\/p>\n<p>: 3<sup>2k<\/sup>\u00a0= 8\u03bb \u2013 7 \u2026 (i)<\/p>\n<p>We have to prove,<\/p>\n<p>3<sup>2(k + 1)<\/sup>\u00a0+ 7 is divisible by 8<\/p>\n<p>3<sup>2k + 2<\/sup>\u00a0+ 7 = 8\u03bc<\/p>\n<p>So,<\/p>\n<p>= 3<sup>2(k + 1)<\/sup>\u00a0+ 7<\/p>\n<p>= 3<sup>2k<\/sup>.3<sup>2<\/sup>\u00a0+ 7<\/p>\n<p>= 9.3<sup>2k<\/sup>\u00a0+ 7<\/p>\n<p>= 9.(8\u03bb\u00a0\u2013 7) + 7 by using equation (i)<\/p>\n<p>= 72\u03bb\u00a0\u2013 63 + 7<\/p>\n<p>= 72\u03bb\u00a0\u2013 56<\/p>\n<p>= 8(9\u03bb\u00a0\u2013 7)<\/p>\n<p>= 8\u03bc<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>21. 5<sup>2n + 2<\/sup>\u00a0\u2013 24n \u2013 25 is divisible by 576 for all n\u00a0\u03f5\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): 5<sup>2n + 2<\/sup>\u00a0\u2013 24n \u2013 25 is divisible by 576.<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 5<sup>2.1+2<\/sup>\u00a0\u2013 24.1 \u2013 25<\/p>\n<p>: 625 \u2013 49<\/p>\n<p>: 576<\/p>\n<p>P (n) is true for n = 1. Where P (n) is divisible by 576<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 5<sup>2k + 2<\/sup>\u00a0\u2013 24k \u2013 25 is divisible by 576<\/p>\n<p>: 5<sup>2k + 2<\/sup>\u00a0\u2013 24k \u2013 25 = 576\u03bb \u2026. (i)<\/p>\n<p>We have to prove,<\/p>\n<p>5<sup>2k + 4<\/sup>\u00a0\u2013 24(k + 1) \u2013 25 is divisible by 576<\/p>\n<p>5<sup>(2k + 2) + 2<\/sup>\u00a0\u2013 24(k + 1) \u2013 25 = 576\u03bc<\/p>\n<p>So,<\/p>\n<p>=\u00a05<sup>(2k + 2) + 2<\/sup>\u00a0\u2013 24(k + 1) \u2013 25<\/p>\n<p>= 5<sup>(2k + 2)<\/sup>.5<sup>2<\/sup>\u00a0\u2013 24k \u2013 24\u2013 25<\/p>\n<p>= (576\u03bb\u00a0+ 24k + 25)25 \u2013 24k\u2013 49 by using equation (i)<\/p>\n<p>= 25. 576\u03bb + 576k + 576<\/p>\n<p>= 576(25\u03bb + k + 1)<\/p>\n<p>= 576\u03bc<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>22. 3<sup>2n + 2<\/sup>\u00a0\u2013 8n \u2013 9 is divisible by 8 for all n\u00a0\u03f5\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): 3<sup>2n + 2<\/sup>\u00a0\u2013 8n \u2013 9 is divisible by 8<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 3<sup>2.1 + 2<\/sup>\u00a0\u2013 8.1 \u2013 9<\/p>\n<p>: 81 \u2013 17<\/p>\n<p>: 64<\/p>\n<p>P (n) is true for n = 1. Where P (n) is divisible by 8<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 3<sup>2k + 2<\/sup>\u00a0\u2013 8k \u2013 9 is divisible by 8<\/p>\n<p>: 3<sup>2k + 2<\/sup>\u00a0\u2013 8k \u2013 9 = 8\u03bb \u2026 (i)<\/p>\n<p>We have to prove,<\/p>\n<p>3<sup>2k + 4<\/sup>\u00a0\u2013 8(k + 1) \u2013 9 is divisible by 8<\/p>\n<p>3<sup>(2k + 2) + 2<\/sup>\u00a0\u2013 8(k + 1) \u2013 9 = 8\u03bc<\/p>\n<p>So,<\/p>\n<p>=\u00a03<sup>2(k + 1)<\/sup>.3<sup>2<\/sup>\u00a0\u2013 8(k + 1) \u2013 9<\/p>\n<p>= (8\u03bb\u00a0+ 8k + 9)9 \u2013 8k \u2013 8 \u2013 9<\/p>\n<p>= 72\u03bb\u00a0+ 72k + 81 \u2013 8k \u2013 17 using equation (1)<\/p>\n<p>= 72\u03bb\u00a0+ 64k + 64<\/p>\n<p>= 8(9\u03bb + 8k + 8)<\/p>\n<p>= 8\u03bc<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>23. (ab)<sup>\u00a0n<\/sup>\u00a0= a<sup>n<\/sup>\u00a0b<sup>n<\/sup>\u00a0for all n\u00a0\u03f5\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): (ab)<sup>\u00a0n<\/sup>\u00a0= a<sup>n<\/sup>\u00a0b<sup>n<\/sup><\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): (ab)<sup>\u00a01<\/sup>\u00a0= a<sup>1<\/sup>\u00a0b<sup>1<\/sup><\/p>\n<p>: ab = ab<\/p>\n<p>P (n) is true for n = 1.<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): (ab)<sup>\u00a0k<\/sup>\u00a0= a<sup>k<\/sup>\u00a0b<sup>k<\/sup>\u00a0\u2026 (i)<\/p>\n<p>We have to prove,<\/p>\n<p>(ab)\u00a0<sup>k + 1\u00a0<\/sup>= a<sup>k + 1<\/sup>.b<sup>k + 1<\/sup><\/p>\n<p>So,<\/p>\n<p>= (ab)<sup>\u00a0k + 1<\/sup><\/p>\n<p>= (ab)<sup>\u00a0k<\/sup>\u00a0(ab)<\/p>\n<p>= (a<sup>k\u00a0<\/sup>b<sup>k<\/sup>) (ab) using equation (1)<\/p>\n<p>= (a<sup>k + 1<\/sup>) (b<sup>k + 1<\/sup>)<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>24. n (n + 1) (n + 5) is a multiple of 3 for all n\u00a0\u03f5\u00a0N.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): n (n + 1) (n + 5) is a multiple of 3<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 1 (1 + 1) (1 + 5)<\/p>\n<p>: 2 \u00d7 6<\/p>\n<p>: 12<\/p>\n<p>P (n) is true for n = 1. Where P (n) is a multiple of 3<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): k (k + 1) (k + 5) is a multiple of 3<\/p>\n<p>: k(k + 1) (k + 5) = 3\u03bb \u2026 (i)<\/p>\n<p>We have to prove,<\/p>\n<p>(k + 1)[(k + 1) + 1][(k + 1) + 5] is a multiple of 3<\/p>\n<p>(k + 1)[(k + 1) + 1][(k + 1) + 5] = 3\u03bc<\/p>\n<p>So,<\/p>\n<p>= (k + 1) [(k + 1) + 1] [(k + 1) + 5]<\/p>\n<p>= (k + 1) (k + 2) [(k + 1) + 5]<\/p>\n<p>= [k (k + 1) + 2(k + 1)] [(k + 5) + 1]<\/p>\n<p>= k (k + 1) (k + 5) + k(k + 1) + 2(k + 1) (k + 5) + 2(k + 1)<\/p>\n<p>= 3\u03bb + k<sup>2<\/sup>\u00a0+ k + 2(k<sup>2<\/sup>\u00a0+ 6k + 5) + 2k + 2<\/p>\n<p>= 3\u03bb + k<sup>2<\/sup>\u00a0+ k + 2k<sup>2<\/sup>\u00a0+ 12k + 10 + 2k + 2<\/p>\n<p>= 3\u03bb + 3k<sup>2<\/sup>\u00a0+ 15k + 12<\/p>\n<p>= 3(\u03bb + k<sup>2<\/sup>\u00a0+ 5k + 4)<\/p>\n<p>= 3\u03bc<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n \u2208 N.<\/p>\n<p><strong>25. 7<sup>2n<\/sup>\u00a0+ 2<sup>3n \u2013 3<\/sup>. 3n \u2013 1 is divisible by 25 for all n\u00a0\u03f5\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let P (n): 7<sup>2n<\/sup>\u00a0+ 2<sup>3n \u2013 3<\/sup>. 3n \u2013 1 is divisible by 25<\/p>\n<p>Let us check for n = 1,<\/p>\n<p>P (1): 7<sup>2<\/sup>\u00a0+ 2<sup>0<\/sup>.3<sup>0<\/sup><\/p>\n<p>: 49 + 1<\/p>\n<p>: 50<\/p>\n<p>P (n) is true for n = 1. Where P (n) is divisible by 25<\/p>\n<p>Now, let us check that P (n) is true for n = k and prove that P (k + 1) is true.<\/p>\n<p>P (k): 7<sup>2k<\/sup>\u00a0+ 2<sup>3k \u2013 3<\/sup>. 3k \u2013 1 is divisible by 25<\/p>\n<p>: 7<sup>2k<\/sup>\u00a0+ 2<sup>3k \u2013 3<\/sup>. 3<sup>k \u2013 1<\/sup>\u00a0= 25\u03bb \u2026 (i)<\/p>\n<p>We have to prove that:<\/p>\n<p>7<sup>2k + 1<\/sup>\u00a0+ 2<sup>3k<\/sup>. 3<sup>k<\/sup>\u00a0is divisible by 25<\/p>\n<p>7<sup>2k + 2<\/sup>\u00a0+ 2<sup>3k<\/sup>. 3<sup>k<\/sup>\u00a0= 25\u03bc<\/p>\n<p>So,<\/p>\n<p>= 7<sup>2(k + 1)<\/sup>\u00a0+ 2<sup>3k<\/sup>. 3<sup>k<\/sup><\/p>\n<p>= 7<sup>2k<\/sup>.7<sup>1<\/sup>\u00a0+ 2<sup>3k<\/sup>. 3<sup>k<\/sup><\/p>\n<p>= (25\u03bb\u00a0\u2013 2<sup>3k \u2013 3<\/sup>. 3<sup>k \u2013 1<\/sup>) 49 + 2<sup>3k<\/sup>. 3k by using equation (i)<\/p>\n<p>= 25\u03bb. 49 \u2013 2<sup>3k<\/sup>\/8. 3<sup>k<\/sup>\/3. 49 + 2<sup>3k<\/sup>. 3<sup>k<\/sup><\/p>\n<p>= 24\u00d725\u00d749\u03bb\u00a0\u2013 2<sup>3k\u00a0<\/sup>. 3<sup>k\u00a0<\/sup>. 49 + 24 . 2<sup>3k<\/sup>.3<sup>k<\/sup><\/p>\n<p>= 24\u00d725\u00d749\u03bb\u00a0\u2013 25 . 2<sup>3k<\/sup>. 3<sup>k<\/sup><\/p>\n<p>= 25(24 . 49\u03bb\u00a0\u2013 2<sup>3k<\/sup>. 3<sup>k<\/sup>)<\/p>\n<p>= 25\u03bc<\/p>\n<p>P (n) is true for n = k + 1<\/p>\n<p>Hence, P (n) is true for all n\u00a0\u2208\u00a0N.<\/p>\n<p>This is the complete blog on the RD Sharma Solutions For Class 11 Maths Chapter 12 Exercise 12.2. To Know more about the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 11 Maths exam, as in the comments.\u00a0<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-solutions-for-class-11-maths-chapter-12-exercise-122\"><\/span>FAQs on RD Sharma Solutions For Class 11 Maths Chapter 12 Exercise 12.2<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-1630491490331\" 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-for-class-11-maths-chapter-12-exercise-122\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 12 Exercise 12.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630491530062\" 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-12-ex-122\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 12 Ex 12.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630491545439\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-12-exercise-122\"><\/span>How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 12 Exercise 12.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 50 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 12 Exercise 12.2: Acing your Class 11 Maths exams is made easier, all thanks to the RD Sharma Solutions Class 11 Maths. You can clear your doubts and strengthen you basics with this ultimate guide. Download the Free PDF of RD Sharma Solutions Class 11 Maths Chapter 12 &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 12 Exercise 12.2 (Updated for 2023-24)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-12-exercise-12-2\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 12 Exercise 12.2 (Updated for 2023-24)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122651,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67944"}],"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=67944"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67944\/revisions"}],"predecessor-version":[{"id":515109,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67944\/revisions\/515109"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122651"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67944"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67944"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67944"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}