Question

show that for any natural number n,3^2n-8n-1 is divisible by 64

Answer 1

Answer:

Step-by-step explanation:

Let f(n) = 3^(2n+2) - 8n - 9

Express f(n+1) in terms of f(n).

f(n+1) = 3^(2(n+1)+2) - 8(n+1) - 9

f(n+1) = 3^(2n+2+2) - 8n - 8 - 9

f(n+1) = 3^(2n+4) - 8n - 17

f(n+1) = 9(3^(2n+2)) - 8n - 17

f(n+1) = 9(3^(2n+2)) - 72n + 64n - 17

f(n+1) = 9(3^(2n+2) - 8n) + 64n - 17

f(n+1) = 9(3^(2n+2) - 8n) + 64n + 64 - 81

f(n+1) = 9(3^(2n+2) - 8n - 9) + 64n + 64

f(n+1) = 9f(n) + 64n + 64

f(n+1) = 64(9f(n)/64 + n + 1)

So if f(n) is divisible by 64, and n is an integer, f(n+1) is also divisible by 64.

Evaluate f(1):

f(1) = 3^(2x1+2) - 8x1 - 9

f(1) = 3^(2+2) - 8 - 9

f(1) = 3^4 - 17

f(1) = 81 - 17

f(1) = 64

f(1) is divisible by 64. Therefore f(n) is divisible by 64 for all integers n greater than or equal to 1.