Can someone point me in the right direction of solving the following problem:

Prove that for any postive integer n, the value of the expression 3^(2n+2)-8n-9 is divisible by 64.

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- Apr 12th 2005, 06:41 PMTatanka[SOLVED] Proof divisible by 64
Can someone point me in the right direction of solving the following problem:

Prove that for any postive integer n, the value of the expression 3^(2n+2)-8n-9 is divisible by 64. - Apr 19th 2005, 04:32 AMhpeQuote:

Originally Posted by**Tatanka**

Set x_n = 3^(2n+2) -8n -9.

Solve for the power term on the right:

x_n + 8n + 9 = 3^(2n+2)

Now look at the term for the next n:

x_n+1 = 3^(2n+4) -8(n+1) - 9

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

= 9*(x_n + 8n + 9) -8n - 17

= ...

Use this to prove that if x_n is divisible by 64, x_n+1 is divisible by 64 as well.

You still need to start the induction for some n:-) - Apr 22nd 2005, 05:12 AMvmsQuote:

Originally Posted by**Tatanka**

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

expanding it by using binomial theorem,

1+(n+1)8+C(n+1, 2)8^2+C(n+1, 3)8^3+C(n+1, 4)8^4....+C(n+1, n+1)8^n+1-8n-9

first two terms get cancel with the last two terms and

then each term contains 8^2 common ...

hence the number remained is always divisible by 64