{"id":13274,"date":"2019-12-19T12:04:19","date_gmt":"2019-12-19T06:34:19","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=13274"},"modified":"2021-07-02T15:35:51","modified_gmt":"2021-07-02T10:05:51","slug":"1st-puc-maths-question-bank-chapter-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-maths-question-bank-chapter-4\/","title":{"rendered":"1st PUC Maths Question Bank Chapter 4 Principle of Mathematical Induction"},"content":{"rendered":"

You can Download Chapter 4 Principle of Mathematical Induction Questions and Answers, Notes, 1st PUC Maths Question Bank with Answers<\/a>, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Maths Question Bank Chapter 4 Principle of Mathematical Induction<\/h2>\n

Question 1.
\nUsing the principle of mathematical induction, prove that
\n\\(\\text { (1) } 1+2+3+\\cdots+n=\\frac{n(n+1)}{2}\\), for all n\u2208N
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k)\u21d2\u00a0P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\n1 + 3 + 5 + 7 + …….. + (2n – 1) = n2<\/sup>, for all n\u2208N
\nAnswer:
\nLet P(n):1 + 3 + 5 + 7+……….. +(2n-1) = n2
\n<\/sup>For n = 1, LHS = 1, RHS = 12<\/sup> = 1
\n\u2234 LHS = RHS
\n\u2234 F(1) is true.
\nLet upwards assume P(k) is true for some k\u2208N
\ni.e., 1 + 3 + 5 + …. + (2K- 1) = kz
\n<\/sup>Adding (k + 1)th<\/sup> term = 2K + 1, on both sides, we get,
\n1 + 3 + 5 +……. +(2k -1)(2k + l) = k2<\/sup> + 2k + 1
\n= (k +1)2
\n<\/sup>which is P(k +1)
\nThus, P(k)\u21d2\u00a0P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 3.
\n\\(1^{2}+2^{2}+3^{2}+\\ldots \\ldots+n^{2}=\\frac{n(n+1)(2 n+1)}{6}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\n\\(1^{3}+2^{3}+3^{3}+\\ldots \\ldots+n^{3}=\\left[\\frac{n(n+1)}{2}\\right]^{2} \\) for all n\u2208N
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 5.
\n\\(1 \\cdot 2+2 \\cdot 3+\\dots+n(n+1)=\\frac{n(n+1)(n+2)}{3}\\) for all n\u2208N
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 6.
\n\\( \\begin{aligned} &1 \\cdot 2 \\cdot 3+2 \\cdot 3 \\cdot 4+\\dots+n(n+1)(n+2)&=\\frac{n(n+1)(n+2)(n+3)}{4},
\n\\text { for all } n \\in N \\end{aligned}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 7.
\n\\( \\cfrac{1}{1 \\cdot 2}+\\cfrac{1}{2 \\cdot 3}+\\cfrac{1}{3 \\cdot 4}+\\dots+\\cfrac{1}{n(n+1)}=\\cfrac{n}{n+1}\\) for all n\u2208N
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 8.
\n\\(\\begin{aligned} &\\cfrac{1}{1 \\cdot 2 \\cdot 3}+\\cfrac{1}{2 \\cdot 3 \\cdot 4}+\\cfrac{1}{3 \\cdot 4 \\cdot 5}+\\dots+\\cfrac{1}{n(n+1)(n+2)}&=\\cfrac{n(n+3)}{4(n+1)(n+2)}, \\text { for all } n \\in N \\end{aligned}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\n\\( \\begin{aligned}&\\frac{1}{2 \\cdot 5}+\\frac{1}{5 \\cdot 8}+\\frac{1}{8 \\cdot 11}+\\cdots \\cdot+\\\\ &\\frac{1}{(3 n-1)(3 n+2)}=\\frac{n}{6 n+4} \\end{aligned}\\)
\nAnswer:
\n\"1st
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 10.
\n\\(\\begin{aligned} &\\cfrac{1}{1 \\cdot 4}+\\cfrac{1}{4 \\cdot 7}+\\cfrac{1}{7 \\cdot 10}+\\cdots+\\\\ &\\cfrac{1}{(3 n-2)(3 n+1)}=\\cfrac{n}{3 n+1} \\end{aligned} \\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 11.
\n\\(\\begin{aligned} &\\cfrac{1}{3 \\cdot 5}+\\cfrac{1}{5 \\cdot 7}+\\cfrac{1}{7 \\cdot 9}+\\cdots+\\\\ &\\cfrac{1}{(2 n+1)(2 n+3)}=\\cfrac{n}{3(2 n+3)} \\end{aligned}\\)
\nAnswer:
\n\"1st<\/p>\n

\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 12.
\n\\(\\begin{aligned} &1+\\cfrac{1}{1+2}+\\cfrac{1}{1+2+3}+\\dots+\\\\ &\\cfrac{1}{1+2+3+\\dots+n}=\\cfrac{2 n}{n+1} \\end{aligned} \\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\n\\(\\begin{aligned}&1 \\cdot 3+3 \\cdot 5+5 \\cdot 7+\\dots+(2 n-1)(2 n+1)&=\\cfrac{n\\left(4 n^{2}+6 n-1\\right)}{3}, \\text { for all } n \\in N \\end{aligned}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 14.
\n\\(1+3+3^{2}+\\cdots+3^{n-1}=\\cfrac{3^{n}-1}{2} \\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 15.
\n\\(\\cfrac{1}{2}+\\cfrac{1}{4}+\\cfrac{1}{8}+\\dots+\\cfrac{1}{2^{n}}=1-\\cfrac{1}{2^{n}} \\)
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 16.
\n\\( a+a r+a r^{2}+\\cdots+a r^{n-1}=\\cfrac{a\\left(r^{n}-1\\right)}{r-1}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 17.
\n\\( 1 \\cdot 2+2 \\cdot 2^{2}+3 \\cdot 2^{3}+\\cdots+n 2^{n}=(n-1) 2^{n+1}+2\\)
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 18.
\n\\( \\begin{aligned} &1 \\cdot 3+2 \\cdot 3^{2}+3 \\cdot 3^{2}+\\cdots+n \\cdot 3^{n}&=\\cfrac{(2 n-1) 3^{n+1}+3}{4}, \\text { for all } n \\in N \\end{aligned} \\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 19.
\n\\(\\left(1+\\frac{1}{1}\\right)\\left(1+\\frac{1}{2}\\right)\\left(1+\\frac{1}{3}\\right) \\cdots\\left(1+\\frac{1}{n}\\right)=n+1\\) for all n\u2208N
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 20.
\n\\( \\left(1+\\cfrac{3}{1}\\right)\\left(1+\\cfrac{5}{4}\\right)\\left(1+\\cfrac{7}{9}\\right) \\cdots\\left(1+\\cfrac{2 n+1}{n^{2}}\\right)=(n+1)^{2}\\)
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 21.
\n\\(1^{2}+3^{2}+5^{2}+\\cdots \\cdot \\cdot+(2 n-1)^{2}=\\cfrac{n(2 n-1)(2 n+1)}{3}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 22.
\nProve that 2n<\/sup>> n for all positives integers
\nAnswer:
\n\"1st
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 23.
\n\\( 1+2+3 \\cdots+n<\\frac{1}{8}(2 n+1)^{2}\\)
\nAnswer:
\n\"1st
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 24.
\n\\(1^{2}+2^{2}+3^{2}+\\cdots+n^{2}>\\frac{n^{3}}{3}, n \\in N \\)
\nAnswer:
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 25.
\n(2n + 7) < (n + 3)2
\n<\/sup>Answer:
\nLet P(n): (2n + 7) < (n + 3)2
\n<\/sup>For = 1, (2 + 7)<(1+ 3)2<\/sup> \u21d29<16 is true.
\n\u2234 P(1) is true.
\nLet us assume P(k) is true for some k\u2208 N
\ni.e., (2k+ 1) < (k + 3)2<\/sup>
\nConsider 2 (k + 1) + 7 = 2k + 2 + 7
\n= (2k + 7) + 2
\n< (k: + 3)2<\/sup> + 2 = k2<\/sup> + 6k + 9 + 2
\n<(k + 4)2
\n<\/sup>\u2234 P(k +1) is true.
\nHence, by mathematical induction, P(n) is true for all n\u2208N<\/p>\n

Question 26.
\n72n<\/sup> – 32n<\/sup> is divisible by 4.
\nAnswer:
\nLet P(n): 7n<\/sup> – 3n<\/sup>\u00a0is divisible by 4.
\nFor n = 1, P(1): 71<\/sup> – 31<\/sup> = 4 which is divisible by 4.
\n\u2234 P(1) is true.
\nLet us assume P(k) is true for some k\u2208N
\ni.e., 72n<\/sup> – 32n <\/sup>is divisible by 4.
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 27.
\nx2n<\/sup> – y2n<\/sup> is divisible by x+y.
\nAnswer:
\nLet P(n): x2n<\/sup> – y2n<\/sup> is divisible by x + y
\nFor n = 1, P(1): x2<\/sup> – y2<\/sup> is divisible by x + y, which is true.
\n\u2234 P(1) is true.
\nLet us assume P(k) is true for some k\u2208 N
\ni.e., x2k<\/sup> – y2k<\/sup> is divisible by x + y
\nLet x2k<\/sup> – y2k<\/sup> = (x + y)d, where d\u2208N
\n\"1st
\nwhich is P(k +1)
\nThus, P(k) \u21d2 P(k +1).
\nHence, by mathematical induction P(n) is true for all n\u2208N<\/p>\n

Question 28.
\n102n<\/sup>-1<\/sup> +1 is divisible by
\nAnswer:
\nLet P(n): 102n-<\/sup>1<\/sup> +1, is divisible by 11.
\nFor n = 1, P(1):102-1<\/sup> +1, is divisible by 11, which is true.
\nP(1) is true.
\nLet us assume P(k) is true for some k\u2208N.
\ni.e., 102n<\/sup>-1<\/sup>+1 is divisible by 11.
\n\u21d2102<\/sup>-1<\/sup> +1 = 1 W, d\u2208N.
\nConsider 102(k+1)-<\/sup>1<\/sup>+1 = 102k<\/sup>+1<\/sup> +1
\n= 102k<\/sup>-1 <\/sup>102<\/sup>+1 = (11d-1)102<\/sup>+1
\n= 11d(102<\/sup>)-100 + 1 = 11(100d)-99
\n= 11(100d-9)
\n\u2234 102(k+1)-1<\/sup> +1 is divisible by 11.
\nThus, P(k) \u21d2 P(k +1)
\nHence, by mathematical induction, P(n) is true for all n\u2208N<\/p>\n

Question 29.
\n32n<\/sup>+2<\/sup> -8n-9 is divisible by 8.
\nAnswer:
\nLet P(n): 32n<\/sup>+2<\/sup> -8n-9\u00a0is divisible by 8.
\nFor n = 1, P(1): 32+2<\/sup> – 8(2) – 9 = 64 is divisible by 8, which is true.
\nP( 1) is true.
\nLet us assume P(k) is true for some k\u2208N
\ni.e., 32k+<\/sup>2<\/sup> -8k -9 is divisible by 8.
\nLet 32k+2<\/sup>-8k-9 = 8d, d\u2208N ………………(1)
\nConsider 32(k<\/sup>+1)+2<\/sup> – 8(k +1) – 9 = 32k<\/sup>+4<\/sup> – 8k – 8 – 9
\n= 32k+2<\/sup> -32<\/sup>-8(k-1) = (8d + 8k + 9)9-8k-17 using (1)
\n= (8d + 8k)9 + 81-8k-17
\n= 8(d + k)9 + 64-8k = 8[9(d + k) + 8-k]
\nwhich is divisible by 8.
\nThus, P(k) \u21d2 P(k +1)
\nHence, by mathematical induction, P(n) is true for all n\u2208N<\/p>\n

\"KSEEB<\/p>\n

Question 30.
\n2.7n<\/sup> + 3.5n<\/sup> – 5 is divisible by 24 for all n\u2208N
\nAnswer:
\nLet P(n): 2.7n<\/sup> + 3.5n<\/sup> – 5 is divisible by 24.
\nFor n = 1; P(1): 2.7n<\/sup> + 3.5n<\/sup> – 5\u00a0= 24 is divisible by 24.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ‘
\n\u2234 P(1) is true.
\nLet us assume, P(k) is true for some k\u2208N
\ni.e P(n): 2.7k<\/sup> + 3.5k<\/sup> – 5 is divisible by 24.
\nLet 2.7k<\/sup> + 3.5k<\/sup> – 5=24d,where d\u2208N
\n\"1st
\nThus, P(k)\u21d2P(k+ l)
\nHence, by mathematical induction, P(n) is true for all n\u2208N<\/p>\n

Question 31.
\n41n<\/sup> -14n<\/sup> \u00a0is a multiple of 27.
\nAnswer:
\nLet P(n): 41n<\/sup> -14n<\/sup> is a multiple of 27.
\nFor n = 1, P(1): 41′-14′ = 47 is a multiple of 27.
\n\u2234 P(1) is true.
\nLet us assume P(k) is true for some k\u2208N
\n\"1st
\nThus, P(k)\u21d2P(k+ 1)
\nHence, by mathematical induction, P(n) is true for all n\u2208N<\/p>\n

Question 32.
\nn(n +1)(n + 5) is a multiple of 3.
\nAnswer:
\nLet P(n): n(n + 1)(n + 5) is a multiple of 3.
\nFor n = 1, P(1): 1(1 +1)(1 + 5) = 12 is a multiple of 3.
\n\u2234 P(1) is true.
\nLet us assume P(k) is true for some n\u2208N
\ni.e., P(k):k(k + 1)(k + 5) is a multiple of 3.
\nLet k(k + 1)(k + 5) = 3d
\nConsider (k + 1)(k + 2)(k + 6)
\n= (k + 2)(k + 1)(k + 5 + 1)
\n= k(k + 1)(k + 5 + 1) + 2(k + 1)(k + 5 + 1)
\n= k(k +1 )(k + 5) + k(k +1) + 2(k +1 )(k + 6)<\/p>\n

Question 33.
\n\\((a b)^{n}=a^{n} b^{n}\\)
\nAnswer:
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 34.
\n(1 + x)n<\/sup> \u2265 (1 + nx)for all natural number n, where x>-l.
\nAnswer:
\nLet P(n): (1 + x)n<\/sup> \u2265 (1 + nx), for x >-1
\nFor n = 1; P(1): (1 + x)1<\/sup> \u2265 (1 + x) which is true.
\nLet us assume P(k) is true for some k\u2208N
\ni.e., (1 + x)k<\/sup> \u2265 (1 + kx)
\nConsider (1 + x)k+1<\/sup> = (1 + x)k<\/sup> (1 + x)
\n\u2265 (1 + kx)(1 + x)
\n= (l + kx + x + kx2<\/sup>)
\n\u2265(1 + (k + 1)x)
\n\u2235 kx2<\/sup>\u22650.
\nwhich is P(k + 1)
\nThus, P(k) \u21d2 P(k +1)
\nHence, by mathematical induction, P(n) is true for all n\u2208N<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Chapter 4 Principle of Mathematical Induction Questions and Answers, Notes, 1st PUC Maths Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Maths Question Bank Chapter 4 Principle of Mathematical Induction Question 1. Using the principle of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13274"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=13274"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/13274\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=13274"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=13274"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=13274"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}