Principle of Mahtematical Induction

Answer :

a_n = 2⁽²⁾ⁿ + 1                   for     n > 1

                   ∴ For n = 2, a₂ = 2⁴ + 1 = 17 = 10 + 7

                      Let           a_k = 2^(2)k + 1 = 10m + 7 be true where k > 1, m ∈ n       (1)

                      Now                  a_k+1 = 2^(2)(k+1) + 1 = (2^(2)k)² + 1 = (10m + 7 – 1)² + 1---from (1)

                                      ∴ a_k+1 = (10m + 6)² + 1

                                                = 10m ² + 120m + 36 + 1

                                                = 10 (m² + 12m + 3) + 7

                                      ∴ Digit of 1's in place of a_n is 7

Answer :

P(n):  [4ⁿ / (n + 1)] < [(2n)! / (n!)²], n ∈ N

                   P(1) = (4/2) < (2! / 1) ------- not true

                   n < 0 is not possible

                   P(n) is not true for all n > 0

                   P(2) = (16 / 3) < (6! / 2!) which is true hence n ≥ 2, n ∈ N  

Answer :

P(n): cos θ cos 2θ cos 4θ….cos (2^n–1θ)

                   P(1): cos θ                                                                               (1)

          Consider option   (a) for n = 1, we get [(sin 2θ) / (2 sin θ)] = cos θ

                                      (b) for n = 1, we get [(cos 2θ) / (2 sin θ)]

                                      (c) for n = 1, we get [(sin 2θ) / (sin θ)] = 2 cos θ

          ∴ From (1), only option (a) satisfies given condition

                   Hence answer is [(sin 2ⁿθ) / (2ⁿ sin θ)]

Answer :

P(n): [1 / (1 ∙ 2 ∙ 3)] + [1 / (2 ∙ 3 ∙ 4)] + … [1 / {n (n + 1) (n + 2)}]

                   For n = 1, we get P(1) = [1 / (1 ∙ 2 ∙ 3)] = (1/6)

                   Consider options given for n = 1,

                             (a) P(1) = [(1 × 2) / (4 × 3 × 4)] = (1 / 24)

                             (b) P(1) = [(1 × 4) / (4 × 2 × 3)] = (1/6)

                             (c) P(1) = [(1 × 3) / (4 × 2 × 4)] = (3 / 32)

Answer :

P(n): 5²ⁿ⁺¹ + 3ⁿ⁺² ∙ 2^n–1

                   P(1): 5³ + 3³ ∙ 2° = 125 + 27 = 152 = 19 × 8

                   Let     P(k): 5^2k+1 + 3^k+2 ∙ 2^k–1 = 19 m, m ∈ N                                     (1)

                   ∴       P(k + 1) = 5^2k+3 + 3^k+3 ∙ 2^k = 5^2k+1 ∙ 5² + 3 ∙ 3^k+2 ∙ 2 ∙ 2^k–1

                                                                   = 25 ∙ 5^2k+1 + 6 ∙ 3^k+2 ∙ 2^k–1

                   From (1) we get

                           P(k + 1)        = 25 (19m – 3^k+2 ∙ 2^k–1) + 6 ∙ 3^k+2 ∙ 2^k–1

                                      = 25 × 19 m – (19 ∙ 3^k+2 ∙ 2^k+1)

                                      = 19 (25 m – 3^k+2 ∙ 2^k–1)

                                      = 19 m¹

                   Hence divisible by 19.