Question

10ⁿ + 3(4^{n + 2}) + 5 is divisible by (for all n∈N)

(a) 5 

(b) 7 

(c) 9 

(d) 13

Answer 1

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.