12n + 25n−1 is divisible by 12.

Correct! Wrong!

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P(n) : 12n + 25n−1is divisible by 13 We verify for n = 1 P(1) : 121 + 251−1 is divisible by 13 121 + 251−1 = 12 +1 = 13, which is divisible by 13 ⇒P(1) is true We assume P(k) to be true P(k): 12k + 25k−1 is divisible by 13 We will prove that P(k + 1) is true P(k + 1) : 12k+1 + 25kis divisible by 13 Proof of P(k + 1): 12k+1 + 25k = 12k(13 − 1) + 25k−1 (26 − 1) = 13(12k + 2.25k−1) − (12k + 25k−1) The first term is divisible by 13 and because of P(K) the second is also divisible by 13 ⇒12k+1 +25k is divisible by 13 ⇒P(k + 1) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

102n−1 + 1 is divisible by 11.

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P(n): 102n−1 + 1 is divisible by 11. Let n = 1 P(1): 102−1+ 1 is divisible by 11 102−1 + 1 = 11, which is divisible by 11 ⇒P(1) is true We assume P(k) to be true P(k) : 102k − 1 + 1 is divisible by 11 We will prove that P(k + 1) is true P(k + 1): 102k + 1 + 1 is divisible by 11 Proof of P(k + 1): 102k + 1 + 1 = 102k−1 . 102+ 1 =102k−1 (99 + 1) + 1 = 99. 102k−1 + (102k−1 + 1) The first term is divisible by 11 and the second is also divisible by 11 (From P(k)) ⇒102k+1 + 1 is divisible by 11 ⇒P(k + 1) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

4n + 15n − 1 is divisible by 9.

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P(n): 4n + 15n − 1 is divisible by 9 We verify for n = 1 P(1): 41 + 15(1) − 1 is divisible by 9 41 + 15(1) − 1 = 18 which is divisible by 9 ⇒P(1) is true We assume P(k) to be true P(k): 4k + 15k − 1 is divisible by 9 We will prove that P(k + 1) is true P(k + 1): 4k+1 +15(k + 1) − 1 is divisible by 9 Proof of P(k + 1): 4k+1 + 15(k + 1) − 1 = 4k. 4 + 15k + 14 4k(9 − 5) + (90k − 75k) + (9 + 5) = 9[4k + 10k + 1] − 5[4k+ 15k − 1] The first term is divisible by 9 and because of P(k), the second term is also divisible by 9 ⇒ 4k+1 + 15k + 14 is divisible by 9 ⇒P(k + 1) is true Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

n2 + n is divisible by ____.

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P(n): n2 + n Let n = 1 P(1): 12 + 1 = 2, which is divisible by 2 We assume P(k) to be true P(k) : k2 + k is divisible by 2 We will prove that P(k + 1) is true P(k + 1): (k + 1)2+ (k + 1) = (k + 2) (k + 1) = k(k + 1) + 2(k + 1) The first term is divisible by 2 [from P(k)] and the second is also divisible by 2. ⇒ (k + 1)2+ (k + 1) is divisible by 2. ⇒ P(k + 1) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

n3 + (n + 1)3 + (n + 2)3is divisible by 9.

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P(n): n3 + (n + 1)3 + (n + 2)3 Let n = 1 P(1): 13 + (1 + 1)3 + (1 + 2)3 13 + 23 + 33 = 1 + 8 + 27 = 36, which is divisible by 9 We assume P(k) to be true P(k) : k3 + (k + 1)3 + (k + 2)3is divisible by 9 We will prove that P(k + 1) is true P(k + 1): (k + 1)3 + (k + 2)3 + (k + 3)3is divisible by 9 Proof of P(k + 1): (k + 1)3 + (k + 2)3 + (k + 3)3 =( k + 1)3 + (k + 2)3 + k3 + 9k2+ 27k + 27 =[ k3 + (k + 1)3 + (k + 2)3] + 9[k2 + 3k + 3] The first term is divisible by 9 [From P(k) and the second is also divisible by 9] ⇒(k + 1)3 + (k + 2)3 + (k + 3)3is divisible by 9 ⇒P(k + 1) is true Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

xn − yn is divisible by x + y when n is even.

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P(n): xn − yn is divisible by x + y when n is even We verify for n = 2 P(x) : x2 − y2 is divisible by x + y x2 − y2 = (x − y) (x + y), which is divisible by x + y ⇒P(2) is true We assume P(k) to be true P(k): xk − yk is divisible by x + y We will prove that P(k + 2) is true P(k + 2): xk+2 − yk+2 is divisible by x + y Proof of P(k + 2): xk+2 − yk+2xk+2 − yk+2 = xk+2 − x2yk + x2yk − yk+2 = x2(xk− yk) + yk(x2 − y2) = x2(xk − yk) + yk (x − y) (x + y) Because of P(k) the first term is divisible by x + y, the second term is also divisible by x + y ⇒ xk+2 − yk+2 is divisible by x + y ⇒P(k + 1) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all even values of n.

41n − 14nis a multiple of 27.

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P(n): 41n − 14nis a multiple of 27 P(1) : 41 − 14 = 27 is a multiple of 27 ⇒P(1) is true We assume P(k) to be true P(k) : 41k − 14k is multiple of 27 We will prove that P(k + 1) is true P(k + 1): 41k+1 − 14k+1 is a multiple of 27 Proof of P(k + 1): 41k+1 − 14k+1 = 41k . 41 − 14k+1 = 41k (27 + 14) − 14k+1 =27. 41k + 14.41k − 14k+1 =27.41k + 14(41k − 14k) 27.41kis a multiple of 27. Also 14(41k − 14k)is a multiple of 27, from P(k) ⇒P(k + 1) is true Hence, P(n) is true for all n.

2.7n + 3.5n− 5 is divisible by 24.

Correct! Wrong!

P(n) : 2.7n + 3.5n− 5 is divisible by 24. We verify for n = 1 P(1): 2 . 71 + 3.51 − 5 is divisible by 24 2.71 + 3.51− 5 = 14 + 15 − 5 = 24, which is divisible by 24 ⇒ P(1) is true We assume P(k) to be true P(k): 2.7k + 3.5k− 5 is divisible by 24 We will prove that P(k + 1) is true P(k + 1) : 2.7(k+1) + 3.5(k+1)− 5 is divisible by 24 Proof of P(k + 1): 2. 2k+1 + 3.5k+1 – 5 = 2.7k(6 + 1) + 3.5k (4 + 1) – 5 = 12(7k + 5k) + [2.7k + 3.5k− 5] 7k and 5k are odd numbers, therefore (7k + 5k) will be even and 12(7k + 5k)will be divisible by 24, the second term will also be divisible by 24 [From p(k)] ⇒2.7k+1 + 3.5k+1 − 5 is divisible by 24 ⇒p(k + 1) is true Hence, lby the principle of Mathematical Induction, P(n) is true for all n.

23n− 1 is divisible by ____.

Correct! Wrong!

P(n): 23n − 1 Let n = 1 P(1) : 23 – 1 23 – 1 = 8 – 1 = 7, which is divisible by 7 We assume P(k) to be true P(k): 23k− 1 is divisible by 7 We will prove that P(k + 1) is true P(k + 1): 23(k+1) – 1 = 23k . 23– 1 = 23k (7 + 1) – 1 = 7. 23k + (23k − 1) The first term is divisible by 7 and the second is also divisible by 7 ⇒ 23(k+1) − 1 is divisible by 7 ⇒P(k + 1) is true. Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

11n+2 + 122n+1 is divisible by 121.

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P(n): 11n+2 + 122n+1 is divisible by 133. We verify for n = 1 P(1): 111+2 + 122+1 is divisible by 133 113 + 123 = (11 + 12)[112 + 122−11 . 12] =(23) (133) which is divisible by 133 We assume P(k) to be true P(k): 11k+2 + 122k+1 is divisible by 133 We will prove that P(k + 1) is true P(k + 1): 11k+3 + 122k+3 is divisible by 133 Proof of P(k + 1): 11k+3 + 122k+3 = 11.11k+2 + 122 . 122k+1 = 11. 11k+2 + (11 + 133) 122k+1 = 11[11k+2 + 122k+1] + 133.122k+1 The first term is divisible by 133 [From P(k)] and the second is also divisible by 133 ⇒11k+3 + 122k+3 is divisible by 133 P(k + 1) is true Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

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CBSE Class 11th Math 4 - Principle of Mathematical Induction MCQs
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