Thread: [SOLVED] Proof divisible by 64

1. [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.

2. Originally Posted by Tatanka
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.
Use mathematical induction.

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:-)

3. Originally Posted by Tatanka
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.

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

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show that 9n 1-8n-9 is divisible by 64

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