{"id":17456,"date":"2022-05-25T19:00:30","date_gmt":"2022-05-25T13:30:30","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17456"},"modified":"2022-05-17T11:05:57","modified_gmt":"2022-05-17T05:35:57","slug":"mcq-questions-for-class-11-maths-chapter-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-4\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Principle of Mathematical Induction Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 4 Principle of Mathematical Induction Objective Questions.<\/p>\n

Principle of Mathematical Induction Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Principle of Mathematical Induction Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Principle of Mathematical Induction Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Principle of Mathematical Induction Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nFor all n\u2208N, 3n5<\/sup> + 5n\u00b3 + 7n is divisible by
\n(a) 5
\n(b) 15
\n(c) 10
\n(d) 3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 15
\nGiven number = 3n5<\/sup> + 5n\u00b2 + 7n
\nLet n = 1, 2, 3, 4, ……..
\n3n5<\/sup> + 5n\u00b3 + 7n = 3 \u00d7 1\u00b2 + 5 \u00d7 1\u00b3 + 7 \u00d7 1 = 3 + 5 + 7 = 15
\n3n5<\/sup> + 5n\u00b3 + 7n = 3 \u00d7 25<\/sup> + 5 \u00d7 2\u00b3 + 7 \u00d7 2 = 3 \u00d7 32 + 5 \u00d7 8 + 7 \u00d7 2 = 96 + 40 + 14 = 150 = 15 \u00d7 10
\n3n5<\/sup> + 5n\u00b3 + 7n = 3 \u00d7 35<\/sup> + 5 \u00d7 3\u00b3 + 7 \u00d7 3 = 3 \u00d7 243 + 5 \u00d7 27 + 7 \u00d7 3 = 729 + 135 + 21 = 885 = 15 \u00d7 59
\nSince, all these numbers are divisible by 15 for n = 1, 2, 3, …..
\nSo, the given number is divisible by 15<\/p>\n<\/details>\n

\n

Question 2.
\n{1 – (1\/2)}{1 – (1\/3)}{1 – (1\/4)} \u2026…. {1 – 1\/(n + 1)} =
\n(a) 1\/(n + 1) for all n \u2208 N.
\n(b) 1\/(n + 1) for all n \u2208 R
\n(c) n\/(n + 1) for all n \u2208 N.
\n(d) n\/(n + 1) for all n \u2208 R<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 1\/(n + 1) for all n \u2208 N.
\nLet the given statement be P(n). Then,
\nP(n): {1 – (1\/2)}{1 – (1\/3)}{1 – (1\/4)} \u2026…. {1 – 1\/(n + 1)} = 1\/(n + 1).
\nWhen n = 1, LHS = {1 \u2013 (1\/2)} = \u00bd and RHS = 1\/(1 + 1) = \u00bd.
\nTherefore LHS = RHS.
\nThus, P(1) is true.
\nLet P(k) be true. Then,
\nP(k): {1 – (1\/2)}{1 – (1\/3)}{1 – (1\/4)} \u2026…. [1 – {1\/(k + 1)}] = 1\/(k + 1)
\nNow, [{1 – (1\/2)}{1 – (1\/3)}{1 – (1\/4)} \u2026…. [1 – {1\/(k + 1)}] \u2219 [1 \u2013 {1\/(k + 2)}]
\n= [1\/(k + 1)] \u2219 [{(k + 2 ) – 1}\/(k + 2)}]
\n= [1\/(k + 1)] \u2219 [(k + 1)\/(k + 2)]
\n= 1\/(k + 2)
\nTherefore p(k + 1): [{1 – (1\/2)}{1 – (1\/3)}{1 – (1\/4)} \u2026…. [1 – {1\/(k + 1)}] = 1\/(k + 2)
\n\u21d2 P(k + 1) is true, whenever P(k) is true.
\nThus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for all n \u2208 N.<\/p>\n<\/details>\n

\n

Question 3.
\nFor all n \u2208 N, 32n<\/sup> + 7 is divisible by
\n(a) non of these
\n(b) 3
\n(c) 11
\n(d) 8<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 8
\nGiven number = 32n + 7
\nLet n = 1, 2, 3, 4, ……..
\n32n<\/sup> + 7 = 3\u00b2 + 7 = 9 + 7 = 16
\n32n<\/sup> + 7 = 34<\/sup> + 7 = 81 + 7 = 88
\n32n<\/sup> + 7 = 36<\/sup> + 7 = 729 + 7 = 736
\nSince, all these numbers are divisible by 8 for n = 1, 2, 3, …..
\nSo, the given number is divisible by 8<\/p>\n<\/details>\n

\n

Question 4.
\nThe sum of the series 1 + 2 + 3 + 4 + 5 + ………..n is
\n(a) n(n + 1)
\n(b) (n + 1)\/2
\n(c) n\/2
\n(d) n(n + 1)\/2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) n(n + 1)\/2
\nGiven, series is series 1 + 2 + 3 + 4 + 5 + ………..n
\nSum = n(n + 1)\/2<\/p>\n<\/details>\n

\n

Question 5.
\nThe sum of the series 1\u00b2 + 2\u00b2 + 3\u00b2 + ……….. n\u00b2 is
\n(a) n(n + 1) (2n + 1)
\n(b) n(n + 1) (2n + 1)\/2
\n(c) n(n + 1) (2n + 1)\/3
\n(d) n(n + 1) (2n + 1)\/6<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) n(n + 1) (2n + 1)\/6
\nGiven, series is 1\u00b2 + 2\u00b2 + 3\u00b2 + ……….. n\u00b2
\nSum = n(n + 1)(2n + 1)\/6<\/p>\n<\/details>\n

\n

Question 6.
\nFor all positive integers n, the number n(n\u00b2 \u2212 1) is divisible by:
\n(a) 36
\n(b) 24
\n(c) 6
\n(d) 16<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 6
\nGiven,
\nnumber = n(n\u00b2 \u2212 1)
\nLet n = 1, 2, 3, 4….
\nn(n\u00b2 – 1) = 1(1 – 1) = 0
\nn(n\u00b2 – 1) = 2(4 – 1) = 2 \u00d7 3 = 6
\nn(n\u00b2 – 1) = 3(9 – 1) = 3 \u00d7 8 = 24
\nn(n\u00b2 – 1) = 4(16 – 1) = 4 \u00d7 15 = 60
\nSince all these numbers are divisible by 6 for n = 1, 2, 3,……..
\nSo, the given number is divisible 6<\/p>\n<\/details>\n

\n

Question 7.
\nIf n is an odd positive integer, then a\u207f + b\u207f is divisible by :
\n(a) a\u00b2 + b\u00b2
\n(b) a + b
\n(c) a – b
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) a + b
\nGiven number = a\u207f + b\u207f
\nLet n = 1, 3, 5, ……..
\na\u207f + b\u207f = a + b
\na\u207f + b\u207f = a\u00b3 + b\u00b3 = (a + b) \u00d7 (a\u00b2 + b\u00b2 + ab) and so on.
\nSince, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
\nSo, the given number is divisible by (a + b)<\/p>\n<\/details>\n

\n

Question 8.
\nn(n + 1) (n + 5) is a multiple of ____ for all n \u2208 N
\n(a) 2
\n(b) 3
\n(c) 5
\n(d) 7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 3
\nLet P(n): n(n + 1)(n + 5) is a multiple of 3.
\nFor n = 1, the given expression becomes (1 \u00d7 2 \u00d7 6) = 12, which is a multiple of 3.
\nSo, the given statement is true for n = 1, i.e. P(1) is true.
\nLet P(k) be true. Then,
\nP(k): k(k + 1)(k + 5) is a multiple of 3
\n\u21d2 K(k + 1) (k + 5) = 3m for some natural number m, …… (i)
\nNow, (k + 1) (k + 2) (k + 6) = (k + 1) (k + 2)k + 6(k + 1) (k + 2)
\n= k(k + 1) (k + 2) + 6(k + 1) (k + 2)
\n= k(k + 1) (k + 5 – 3) + 6(k + 1) (k + 2)
\n= k(k + 1) (k + 5) – 3k(k + 1) + 6(k + 1) (k + 2)
\n= k(k + 1) (k + 5) + 3(k + 1) (k +4) [on simplification]
\n= 3m + 3(k + 1 ) (k + 4) [using (i)]
\n= 3[m + (k + 1) (k + 4)], which is a multiple of 3
\n\u21d2 P(k + 1) (k + 1 ) (k + 2) (k + 6) is a multiple of 3
\n\u21d2 P(k + 1) is true, whenever P(k) is true.
\nThus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for all n \u2208 N.<\/p>\n<\/details>\n

\n

Question 9.
\nFor any natural number n, 7\u207f – 2\u207f is divisible by
\n(a) 3
\n(b) 4
\n(c) 5
\n(d) 7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 5
\nGiven, 7\u207f – 2\u207f
\nLet n = 1
\n7\u207f – 2\u207f = 71<\/sup> – 21<\/sup> = 7 – 2 = 5
\nwhich is divisible by 5
\nLet n = 2
\n7\u207f – 2\u207f = 72 – 22 = 49 – 4 = 45
\nwhich is divisible by 5
\nLet n = 3
\n7\u207f – 2\u207f = 7\u00b3 – 2\u00b3 = 343 – 8 = 335
\nwhich is divisible by 5
\nHence, for any natural number n, 7\u207f – 2\u207f is divisible by 5<\/p>\n<\/details>\n

\n

Question 10.
\nThe sum of the series 1\u00b3 + 2\u00b3 + 3\u00b3 + ………..n\u00b3 is
\n(a) {(n + 1)\/2}\u00b2
\n(b) {n\/2}\u00b2
\n(c) n(n + 1)\/2
\n(d) {n(n + 1)\/2}\u00b2<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) {n(n + 1)\/2}\u00b2
\nGiven, series is 1\u00b3 + 2\u00b3 + 3\u00b3 + ……….. n\u00b3
\nSum = {n(n + 1)\/2}\u00b2<\/p>\n<\/details>\n

\n

Question 11.
\n(1\u00b2 + 2\u00b2 + \u2026… + n\u00b2) _____ for all values of n \u2208 N
\n(a) = n\u00b3\/3
\n(b) < n\u00b3\/3
\n(c) > n\u00b3\/3
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) > n\u00b3\/3
\nLet P(n): (1\u00b2 + 2\u00b2 + \u2026.. + n\u00b2) > n\u00b3\/3.
\nWhen = 1, LHS = 1\u00b2 = 1 and RHS = 1\u00b3\/3 = 1\/3.
\nSince 1 > 1\/3, it follows that P(1) is true.
\nLet P(k) be true. Then,
\nP(k): (1\u00b2 + 2\u00b2 + \u2026.. + k\u00b2 ) > k\u00b3\/3 …. (i)
\nNow,
\n1\u00b2 + 2\u00b2 + \u2026.. + k\u00b2
\n+ (k + 1)\u00b2
\n= {1\u00b2 + 2\u00b2 + \u2026.. + k\u00b2 + (k + 1)\u00b2
\n> k\u00b3\/3 + (k + 1)\u00b3 [using (i)]
\n= 1\/3 \u2219 (k\u00b3 + 3 + (k + 1)\u00b2) = 1\/3 \u2219 {k\u00b2 + 3k\u00b2 + 6k + 3}
\n= 1\/3[k\u00b3 + 1 + 3k(k + 1) + (3k + 2)]
\n= 1\/3 \u2219 [(k + 1)\u00b3 + (3k + 2)]
\n> 1\/3(k + 1)\u00b3
\nP(k + 1):
\n1\u00b2 + 2\u00b2 + \u2026.. + k\u00b2 + (k + 1)\u00b2
\n> 1\/3 \u2219 (k + 1)\u00b3
\nP(k + 1) is true, whenever P(k) is true.
\nThus P(1) is true and P(k + 1) is true whenever p(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for all n \u2208 N.<\/p>\n<\/details>\n

\n

Question 12.
\n{1\/(3 \u2219 5)} + {1\/(5 \u2219 7)} + {1\/(7 \u2219 9)} + \u2026…. + 1\/{(2n + 1) (2n + 3)} =
\n(a) n\/(2n + 3)
\n(b) n\/{2(2n + 3)}
\n(c) n\/{3(2n + 3)}
\n(d) n\/{4(2n + 3)}<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) n\/{3(2n + 3)}
\nLet the given statement be P(n). Then,
\nP(n): {1\/(3 \u2219 5) + 1\/(5 \u2219 7) + 1\/(7 \u2219 9) + \u2026\u2026. + 1\/{(2n + 1)(2n + 3)} = n\/{3(2n + 3).
\nPutting n = 1 in the given statement, we get
\nand LHS = 1\/(3 \u2219 5) = 1\/15 and RHS = 1\/{3(2 \u00d7 1 + 3)} = 1\/15.
\nLHS = RHS
\nThus, P(1) is true.
\nLet P(k) be true. Then,
\nP(k): {1\/(3 \u2219 5) + 1\/(5 \u2219 7) + 1\/(7 \u2219 9) + \u2026\u2026.. + 1\/{(2k + 1)(2k + 3)} = k\/{3(2k + 3)} \u2026.. (i)
\nNow, 1\/(3 \u2219 5) + 1\/(5 \u2219 7) + ..\u2026\u2026 + 1\/[(2k + 1)(2k + 3)] + 1\/[{2(k + 1) + 1}2(k + 1) + 3
\n= {1\/(3 \u2219 5) + 1\/(5 \u2219 7) + \u2026\u2026. + [1\/(2k + 1)(2k + 3)]} + 1\/{(2k + 3)(2k + 5)}
\n= k\/[3(2k + 3)] + 1\/[2k + 3)(2k + 5)] [using (i)]
\n= {k(2k + 5) + 3}\/{3(2k + 3)(2k + 5)}
\n= (2k\u00b2 + 5k + 3)\/[3(2k + 3)(2k + 5)]
\n= {(k + 1)(2k + 3)}\/{3(2k + 3)(2k + 5)}
\n= (k + 1)\/{3(2k + 5)}
\n= (k + 1)\/[3{2(k + 1) + 3}]
\n= P(k + 1): 1\/(3 \u2219 5) + 1\/(5 \u2219 7) + \u2026\u2026.. + 1\/[2k + 1)(2k + 3)] + 1\/[{2(k + 1) + 1}{2(k + 1) + 3}]
\n= (k + 1)\/{3{2(k + 1) + 3}]
\n\u21d2 P(k + 1) is true, whenever P(k) is true.
\nThus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for n \u2208 N.<\/p>\n<\/details>\n

\n

Question 13.
\nIf n is an odd positive integer, then a\u207f + b\u207f is divisible by :
\n(a) a\u00b2 + b\u00b2
\n(b) a + b
\n(c) a – b
\n(d) none of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) a + b
\nGiven number = a\u207f + b\u207f
\nLet n = 1, 3, 5, ……..
\na\u207f + b\u207f = a + b
\na\u207f + b\u207f = a\u00b3 + b\u00b3 = (a + b) \u00d7 (a\u00b2 + b\u00b2 + ab) and so on.
\nSince, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
\nSo, the given number is divisible by (a + b)<\/p>\n<\/details>\n

\n

Question 14.
\n(2 \u2219 7N<\/sup> + 3 \u2219 5N<\/sup> – 5) is divisible by ……….. for all N \u2208 N
\n(a) 6
\n(b) 12
\n(c) 18
\n(d) 24<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 24
\nLet P(n): (2 \u2219 7\u207f + 3 \u2219 5\u207f – 5) is divisible by 24.
\nFor n = 1, the given expression becomes (2 \u2219 71<\/sup> + 3 \u2219 51<\/sup> – 5) = 24, which is clearly divisible by 24.
\nSo, the given statement is true for n = 1, i.e., P(1) is true.
\nLet P(k) be true. Then,
\nP(k): (2 \u2219 7\u207f + 3 \u2219 5\u207f – 5) is divisible by 24.
\n\u21d2 (2 \u2219 7\u207f + 3 \u2219 5\u207f – 5) = 24m, for m = N
\nNow, (2 \u2219 7\u207f + 3 \u2219 5\u207f – 5)
\n= (2 \u2219 7k<\/sup> \u2219 7 + 3 \u2219 5k<\/sup> \u2219 5 – 5)
\n= 7(2 \u2219 7k<\/sup> + 3 \u2219 5k<\/sup> – 5) = 6 \u2219 5k<\/sup> + 30
\n= (7 \u00d7 24m) – 6(5k<\/sup> – 5)
\n= (24 \u00d7 7m) – 6 \u00d7 4p, where (5k<\/sup> – 5) = 5(5k-1<\/sup> – 1) = 4p
\n[Since (5k-1<\/sup> – 1) is divisible by (5 – 1)]
\n= 24 \u00d7 (7m – p)
\n= 24r, where r = (7m – p) \u2208 N
\n\u21d2 P (k + 1): (2 \u2219 7k<\/sup> + 13 \u2219 5k<\/sup> + 1 – 5) is divisible by 24.
\n\u21d2 P(k + 1) is true, whenever P(k) is true.
\nThus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for all n \u2208 N.<\/p>\n<\/details>\n

\n

Question 15.
\nFor all n\u2208N, 52n<\/sup> \u2212 1 is divisible by
\n(a) 26
\n(b) 24
\n(c) 11
\n(d) 25<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 24
\nGiven number = 52n<\/sup> \u2212 1
\nLet n = 1, 2, 3, 4, ……..
\n52n<\/sup> \u2212 1 = 5\u00b2 \u2212 1 = 25 – 1 = 24
\n52n<\/sup> \u2212 1 = 54<\/sup> – 1 = 625 – 1 = 624 = 24 \u00d7 26
\n52n<\/sup> \u2212 1 = 56<\/sup> – 1 = 15625 – 1 = 15624 = 651 \u00d7 24
\nSince, all these numbers are divisible by 24 for n = 1, 2, 3, …..
\nSo, the given number is divisible by 24<\/p>\n<\/details>\n

\n

Question 16.
\n1 \u2219 2 + 2 \u2219 3 + 3 \u2219 4 + ….. + n(n + 1) =
\n(a) n(n + 1)(n + 2)
\n(b) {n(n + 1)(n + 2)}\/2
\n(c) {n(n + 1)(n + 2)}\/3
\n(d) {n(n + 1)(n + 2)}\/4<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) {n(n + 1)(n + 2)}\/3
\nLet the given statement be P(n). Then,
\nP(n): 1 \u2219 2 + 2 \u2219 3 + 3 \u2219 4 + ….. + n(n + 1) = (1\/3){n(n + 1) (n + 2)}
\nThus, the given statement is true for n = 1, i.e., P(1) is true.
\nLet P(k) be true. Then,
\nP(k): 1 \u2219 2 + 2 \u2219 3 + 3 \u2219 4 + ….. + k(k + 1) = (1\/3){k(k + 1) (k + 2)}.
\nNow, 1 \u2219 2 + 2 \u2219 3 + 3 \u2219 4 +…+ k(k + 1) + (k + 1) (k + 2)
\n= (1 \u2219 2 + 2 \u2219 3 + 3 \u2219 4 + ……. + k(k + 1)) + (k + 1) (k + 2)
\n= (1\/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [using (i)]
\n= (1\/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)
\n= (1\/3){(k + 1) (k + 2)(k + 3)}
\n\u21d2 P(k + 1): 1 \u2219 2 + 2 \u2219 3 + 3 \u2219 4 +……+ (k + 1) (k + 2)
\n= (1\/3){k + 1 )(k + 2) (k +3)}
\n\u21d2 P(k + 1) is true, whenever P(k) is true.
\nThus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for all values of \u2208 N.<\/p>\n<\/details>\n

\n

Question 17.
\n1\/(1 \u2219 2 \u2219 3) + 1\/(2 \u2219 3 \u2219 4) + \u2026\u2026.. + 1\/{n(n + 1)(n + 2)} =
\n(a) {n(n + 3)}\/{4(n + 1)(n + 2)}
\n(b) (n + 3)\/{4(n + 1)(n + 2)}
\n(c) n\/{4(n + 1)(n + 2)}
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) {n(n + 3)}\/{4(n + 1)(n + 2)}
\nLet P (n): 1\/(1 \u2219 2 \u2219 3) + 1\/(2 \u2219 3 \u2219 4) + \u2026\u2026. + 1\/{n(n + 1)(n + 2)} = {n(n + 3)}\/{4(n + 1)(n + 2)}
\nPutting n = 1 in the given statement, we get
\nLHS = 1\/(1 \u2219 2 \u2219 3) = 1\/6 and RHS = {1 \u00d7 (1 + 3)}\/[4 \u00d7 (1 + 1)(1 + 2)] = ( 1 \u00d7 4)\/(4 \u00d7 2 \u00d7 3) = 1\/6.
\nTherefore LHS = RHS.
\nThus, the given statement is true for n = 1, i.e., P(1) is true.
\nLet P(k) be true. Then,
\nP(k): 1\/(1 \u2219 2 \u2219 3) + 1\/(2 \u2219 3 \u2219 4) + \u2026\u2026… + 1\/{k(k + 1) (k + 2)} = {k(k + 3)}\/{4(k + 1) (k + 2)}. \u2026\u2026.(i)
\nNow, 1\/(1 \u2219 2 \u2219 3) + 1\/(2 \u2219 3 \u2219 4) + \u2026\u2026\u2026\u2026.. + 1\/{k(k + 1) (k + 2)} + 1\/{(k + 1) (k + 2) (k + 3)}
\n= [1\/(1 \u2219 2 \u2219 3) + 1\/(2 \u2219 3 \u2219 4) + \u2026\u2026\u2026..\u2026. + 1\/{ k(k + 1) (k + 2}] + 1\/{(k + 1)(k + 2) (k + 3)}
\n= [{k(k + 3)}\/{4(k + 1)(k + 2)} + 1\/{(k + 1)(k + 2)(k + 3)}] [using(i)]
\n= {k(k + 3)\u00b2 + 4}\/{4(k + 1)(k + 2) (k + 3)}
\n= (k\u00b3 + 6k\u00b2 + 9k + 4)\/{4(k + 1) (k + 2) (k + 3)}
\n= {(k + 1) (k + 1) (k + 4)}\/{4 (k + 1) (k + 2) (k + 3)}
\n= {(k + 1) (k + 4)}\/{4(k + 2) (k + 3)
\n\u21d2 P(k + 1): 1\/(1 \u2219 2 \u2219 3) + 1\/(2 \u2219 3 \u2219 4) + \u2026\u2026\u2026.\u2026.. + 1\/{(k + 1) (k + 2) (k + 3)}
\n= {(k + 1) (k + 2)}\/{4(k + 2) (k + 3)}
\n\u21d2 P(k + 1) is true, whenever P(k) is true.
\nThus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
\nHence, by the principle of mathematical induction, P(n) is true for all n \u2208 N.<\/p>\n<\/details>\n

\n

Question 18.
\nFor any natural number n, 7\u207f – 2\u207f is divisible by
\n(a) 3
\n(b) 4
\n(c) 5
\n(d) 7<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 5
\nGiven, 7\u207f – 2\u207f
\nLet n = 1
\n7\u207f – 2\u207f = 71<\/sup> – 21<\/sup> = 7 – 2 = 5
\nwhich is divisible by 5
\nLet n = 2
\n7\u207f – 2\u207f = 7\u00b2 – 2\u00b2 = 49 – 4 = 45
\nwhich is divisible by 5
\nLet n = 3
\n7\u207f – 2\u207f = 7\u00b3 – 2\u00b3 = 343 – 8 = 335
\nwhich is divisible by 5
\nHence, for any natural number n, 7\u207f – 2\u207f is divisible by 5<\/p>\n<\/details>\n

\n

Question 19.
\nThe sum of n terms of the series 1\u00b2 + 3\u00b2 + 5\u00b2 +……… is
\n(a) n(4n\u00b2 – 1)\/3
\n(b) n\u00b2(2n\u00b2 + 1)\/6
\n(c) none of these.
\n(d) n\u00b2(n\u00b2 + 1)\/3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) n(4n\u00b2 – 1)\/3
\nLet S = 1\u00b2 + 3\u00b2 + 5\u00b2 +………(2n – 1)\u00b2
\n\u21d2 S = {1\u00b2 + 2\u00b2 + 3\u00b2 + 4\u00b2 ………(2n – 1)\u00b2 + (2n)\u00b2} – {2\u00b2 + 4\u00b2 + 6\u00b2 +………+ (2n)\u00b2}
\n\u21d2 S = {2n \u00d7 (2n + 1) \u00d7 (4n + 1)}\/6 – {4n \u00d7 (n + 1) \u00d7 (2n + 1)}\/6
\n\u21d2 S = n(4n\u00b2 – 1)\/3<\/p>\n<\/details>\n

\n

Question 20.
\nFor all n \u2208 N, 3n5<\/sup> + 5n\u00b3 + 7n is divisible by:
\n(a) 5
\n(b) 15
\n(c) 10
\n(d) 3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 15
\nGiven number = 3n5<\/sup> + 5n\u00b3 + 7n
\nLet n = 1, 2, 3, 4, ……..
\n3n5<\/sup> + 5n\u00b3 + 7n = 3 \u00d7 1\u00b2 + 5 \u00d7 1\u00b3 + 7 \u00d7 1 = 3 + 5 + 7 = 15
\n3n5<\/sup> + 5n\u00b3 + 7n = 3 \u00d7 25<\/sup> + 5 \u00d7 2\u00b3 + 7 \u00d7 2 = 3 \u00d7 32 + 5 \u00d7 8 + 7 \u00d7 2 = 96 + 40 + 14 = 150 = 15 \u00d7 10
\n3n5<\/sup> + 5n\u00b3 + 7n = 3 \u00d7 35<\/sup> + 5 \u00d7 3\u00b3 + 7 \u00d7 3 = 3 \u00d7 243 + 5 \u00d7 27 + 7 \u00d7 3 = 729 + 135 + 21 = 885 = 15 \u00d7 59
\nSince, all these numbers are divisible by 15 for n = 1, 2, 3, …..
\nSo, the given number is divisible by 15<\/p>\n<\/details>\n

\n

We believe the knowledge shared regarding NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download has been useful to the possible extent. If you have any other queries regarding CBSE Class 11 Maths Principle of Mathematical Induction MCQs Multiple Choice Questions with Answers, feel free to reach us via the comment section and we will guide you with the possible solution.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. …<\/p>\n

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