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

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;

=  3·4ⁿ + 51

= 3·4ⁿ⁺¹ + 51

- = 3·4ⁿ⁺¹ + 51 - (3·4ⁿ + 51) = 3·4ⁿ⁺¹ - 3·4ⁿ

- = 3( 4ⁿ⁺¹ - 4ⁿ) = 3×4ⁿ×(4 - 1) = 9×4ⁿ

∴ - is divisible by 9

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

∴ S₀ is divisible by 9

Since   - is divisible by 9, we have;

- =   - is divisible by 9

Therefore is divisible by 9 and  is divisible by 9 for all positive integers n

Step-by-step explanation:

the f

last graph i believe

step-by-step explanation:

worked it out i believe

the answer is b.

step-by-step explanation:

identity property is where 0+whatever number=whatever number. basically 0+ a number.