# CBSE Class 11th Math 4 – Principle of Mathematical Induction MCQs

#### n^{3} + (n + 1)^{3} + (n + 2)^{3}is divisible by 9.

P(n): n^{3} + (n + 1)^{3} + (n + 2)^{3 } Let n = 1 P(1): 1^{3} + (1 + 1)^{3} + (1 + 2)^{3 } 1^{3} + 2^{3} + 3^{3} = 1 + 8 + 27 = 36, which is divisible by 9 We assume P(k) to be true P(k) : k^{3 }+ (k + 1)^{3} + (k + 2)^{3}is divisible by 9 We will prove that P(k + 1) is true P(k + 1): (k + 1)^{3 }+ (k + 2)^{3} + (k + 3)^{3}is divisible by 9 Proof of P(k + 1): (k + 1)^{3} + (k + 2)^{3} + (k + 3)^{3} =( k + 1)^{3} + (k + 2)^{3} + k^{3 }+ 9k^{2}+ 27k + 27 =[ k^{3} + (k + 1)^{3} + (k + 2)^{3}] + 9[k^{2} + 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)^{3}is divisible by 9 ⇒P(k + 1) is true Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

#### 2.7^{n} + 3.5^{n}− 5 is divisible by 24.

P(n) : 2.7^{n} + 3.5^{n}− 5 is divisible by 24. We verify for n = 1 P(1): 2 . 7^{1} + 3.5^{1 }− 5 is divisible by 24 2.7^{1} + 3.5^{1}− 5 = 14 + 15 − 5 = 24, which is divisible by 24 ⇒ P(1) is true We assume P(k) to be true P(k): 2.7^{k} + 3.5^{k}− 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. 2^{k+1 }+ 3.5^{k+1 }– 5 = 2.7^{k}(6 + 1) + 3.5^{k} (4 + 1) – 5 = 12(7^{k }+ 5^{k}) + [2.7^{k }+ 3.5^{k}− 5] 7^{k} and 5^{k} are odd numbers, therefore (7^{k} + 5^{k}) will be even and 12(7^{k} + 5^{k})will be divisible by 24, the second term will also be divisible by 24 [From p(k)] ⇒2.7^{k+1 }+ 3.5^{k+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.

#### 2^{3n}− 1 is divisible by ____.

P(n): 2^{3}n − 1 Let n = 1 P(1) : 2^{3} – 1 2^{3} – 1 = 8 – 1 = 7, which is divisible by 7 We assume P(k) to be true P(k): 2^{3k}− 1 is divisible by 7 We will prove that P(k + 1) is true P(k + 1): 2^{3(k+1) }– 1 = 2^{3k} . 2^{3}– 1 = 2^{3k} (7 + 1) – 1 = 7. 2^{3k} + (2^{3k} − 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.

#### 41^{n} − 14^{n}is a multiple of 27.

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

#### 12^{n} + 25^{n−1} is divisible by 12.

P(n) : 12^{n} + 25^{n−1}is divisible by 13 We verify for n = 1 P(1) : 12^{1} + 25^{1−1 }is divisible by 13 12^{1} + 25^{1−1 }= 12 +1 = 13, which is divisible by 13 ⇒P(1) is true We assume P(k) to be true P(k): 12k + 25^{k−1 }is divisible by 13 We will prove that P(k + 1) is true P(k + 1) : 12^{k+1} + 25^{k}is divisible by 13 Proof of P(k + 1): 12^{k+1 }+ 25^{k} = 12^{k}(13 − 1) + 25^{k−1 } (26 − 1) = 13(12^{k} + 2.25^{k−1}) − (12^{k} + 25^{k−1}) The first term is divisible by 13 and because of P(K) the second is also divisible by 13 ⇒12^{k+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.

#### n^{2} + n is divisible by ____.

P(n): n^{2} + n Let n = 1 P(1): 1^{2} + 1 = 2, which is divisible by 2 We assume P(k) to be true P(k) : k^{2} + 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.

#### 10^{2n−1 }+ 1 is divisible by 11.

P(n): 10^{2n−1 }+ 1 is divisible by 11. Let n = 1 P(1): 10^{2−1}+ 1 is divisible by 11 10^{2−1 }+ 1 = 11, which is divisible by 11 ⇒P(1) is true We assume P(k) to be true P(k) : 10^{2k − 1 }+ 1 is divisible by 11 We will prove that P(k + 1) is true P(k + 1): 10^{2k + 1 }+ 1 is divisible by 11 Proof of P(k + 1): 10^{2k + 1 }+ 1 = 10^{2k−1 }. 10^{2}+ 1 =10^{2k−1 } (99 + 1) + 1 = 99. 10^{2k−1 }+ (10^{2k−1 }+ 1) The first term is divisible by 11 and the second is also divisible by 11 (From P(k)) ⇒10^{2k+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.

#### 4^{n }+ 15n − 1 is divisible by 9.

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

#### 11^{n+2} + 12^{2n+1 }is divisible by 121.

P(n): 11^{n+2 }+ 12^{2n+1} is divisible by 133. We verify for n = 1 P(1): 11^{1+2 }+ 12^{2+1 }is divisible by 133 11^{3 }+ 12^{3} = (11 + 12)[11^{2} + 12^{2}−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): 11^{k+3 }+ 12^{2k+3 }is divisible by 133 Proof of P(k + 1): 11^{k+3 }+ 12^{2k+3 }= 11.11^{k+2 }+ 12^{2 }. 12^{2k+1} = 11. 11^{k+2 }+ (11 + 133) 12^{2k+1} = 11[11^{k+2 }+ 12^{2k+1}] + 133.12^{2k+1} The first term is divisible by 133 [From P(k)] and the second is also divisible by 133 ⇒11^{k+3 }+ 12^{2k+3 }is divisible by 133 P(k + 1) is true Hence, by the Principle of Mathematical Induction, P(n) is true for all n.

#### x^{n }− y^{n} is divisible by x + y when n is even.

P(n): x^{n }− y^{n } is divisible by x + y when n is even We verify for n = 2 P(x) : x^{2 }− y^{2} is divisible by x + y x^{2} − y^{2} = (x − y) (x + y), which is divisible by x + y ⇒P(2) is true We assume P(k) to be true P(k): x^{k }− y^{k} is divisible by x + y We will prove that P(k + 2) is true P(k + 2): x^{k+2 }− y^{k+2} is divisible by x + y Proof of P(k + 2): x^{k+2 }− y^{k+2xk+2 − yk+2 = xk+2 − x2yk + x2yk − yk+2 } = x^{2}(x^{k}− y^{k}) + y^{k}(x^{2} − y^{2}) = x^{2}(x^{k} − y^{k}) + y^{k} (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 ⇒ x^{k+2 }− y^{k+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.