Question

Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.​

Answer 1

Answer:

To prove that 3·4ⁿ + 51 is divisible by 3 and 9, we have;

3·4ⁿ is divisible by 3 and 51 is divisible by 3

Where we have;

$S_{(n)}$ =  3·4ⁿ + 51

$S_{(n+1)}$ = 3·4ⁿ⁺¹ + 51

$S_{(n+1)}$ - $S_{(n)}$ = 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ

$S_{(n+1)}$ - $S_{(n)}$ = 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ

∴ $S_{(n+1)}$ - $S_{(n)}$ is divisible by 9

Given that we have for S₀ =  3×4⁰ + 51 = 63 = 9×7

∴ S₀ is divisible by 9

Since  $S_{(n+1)}$ - $S_{(n)}$ is divisible by 9, we have;

$S_{(0+1)}$ - $S_{(0)}$ =  $S_{(1)}$ - $S_{(0)}$ is divisible by 9

Therefore $S_{(1)}$ is divisible by 9 and $S_{(n)}$  is divisible by 9 for all positive integers n

Step-by-step explanation: