10n + 3(4n + 2) + 5 is divisible by (for all n∈N)(a) 5 (b) 7 (c) 9

1.	10n + 3(4n + 2) + 5 is divisible by (for all n∈N)(a) 5 (b) 7 (c) 9 (d) 13
Answer» Answer: (C) = 9 For n = 1, 10ⁿ + 3 (4^{n + 2}) + 5 = 10 + 3 × 4³ + 5 = 10 + 192 + 5 = 207 which is divisible by only 9 and none of the other given alternatives. \(\therefore\) We need to prove 10ⁿ+ 3 (4^{n + 2}) + 5 is divisible by 9 ∀ n∈N. Let T(n) be the statement 10ⁿ + 3(4ⁿ⁺²) + 5 is divisible by 9. Basic Step: For n = 1, T(1) holds true as proved above. Induction Step: Assume T(k) to be true, k∈N i.e., 10^k + 3 (4^{k + 2}) + 5 is divisible by 9, i.e., 10^k + 3 (4^{k + 2}) + 5 = 9m, m∈N ....(i) Now, 10^{k + 1} + 3 (4^k+1+2) + 5 = 10^k+1 + 3(4^{k + 3}) + 5 = 10. 10^k + 12. 4^k+2 + 5 = 4(10^k + 3(4^{k + 2})+5) + 6 .10^k – 15 = 4. (9m) + 6 (10^k– 1) – 9 = 4. (9m) + 6. (9x) – 9 (\(\because\) 10k – 1 is always divisible by 9) = 9 (4m + 6x – 1) ⇒ 10^{k + 1} + 3 (4^{(k + 1) + 2}) + 5 is divisible by 9. ⇒ T(k + 1) is true whenever T(k) is true, ∀ k∈N ⇒ 10ⁿ – 3(4^{n + 2}) + 5 is divisible by 9 ∀ k∈N.

10n + 3(4n + 2) + 5 is divisible by (for all n∈N)(a) 5 (b) 7 (c) 9 (d) 13

Answer»

Answer: (C) = 9

For n = 1,

10ⁿ + 3 (4^{n + 2}) + 5 = 10 + 3 × 4³ + 5 = 10 + 192 + 5 = 207 which is divisible by only 9 and none of the other given alternatives.

\(\therefore\) We need to prove 10ⁿ+ 3 (4^{n + 2}) + 5 is divisible by 9 ∀ n∈N.

Let T(n) be the statement

10ⁿ + 3(4ⁿ⁺²) + 5 is divisible by 9.

Basic Step:

For n = 1, T(1) holds true as proved above.

Induction Step:

Assume T(k) to be true, k∈N i.e.,

10^k + 3 (4^{k + 2}) + 5 is divisible by 9, i.e.,

10^k + 3 (4^{k + 2}) + 5 = 9m, m∈N ....(i)

Now, 10^{k + 1} + 3 (4^k+1+2) + 5

= 10^k+1 + 3(4^{k + 3}) + 5

= 10. 10^k + 12. 4^k+2 + 5

= 4(10^k + 3(4^{k + 2})+5) + 6 .10^k – 15

= 4. (9m) + 6 (10^k– 1) – 9

= 4. (9m) + 6. (9x) – 9 (\(\because\) 10k – 1 is always divisible by 9)

= 9 (4m + 6x – 1)

⇒ 10^{k + 1} + 3 (4^{(k + 1) + 2}) + 5 is divisible by 9.

⇒ T(k + 1) is true whenever T(k) is true, ∀ k∈N

⇒ 10ⁿ – 3(4^{n + 2}) + 5 is divisible by 9 ∀ k∈N.