# prove 3^(2n+2)-8n-9 is divisible by 8 for all natural numbers

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### 2 Answers

It has to be shown that 3^(2n+2)-8n-9 is divisible by 8 for all natural values of n.

First, consider the value of 3^(2n+2)-8n-9 for n = 1, it is 3^(2+2)-8-9 = 81 - 17 = 64

Now assume 3^(2n+2)-8n-9 is divisible by 8 for a value of n. With this assumption we test whether 3^(2n+2)-8n-9 is divisible by 8 for n = n+1

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

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

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

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

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

This is clearly divisible by 8 as we have assumed 3^(2n+2) - 8n - 9 is divisible by 8 and 8*(3^(2n+2) - n - 1) has 8 as a factor.

This proves that 3^(2n+2)-8n-9 is divisible by 8 for all natural values of n

Prove 3^(2n+2)-8n-9 is divisible by 8 for all natural numbers. We prove it by method of mathematical induction.

Define statement `P(n):3^(2n+2)-8n-9` is divisible by 8.

P(1): `3^(2xx1+2)-8xx1-9=64/8=8`

Thus P(1) is true. Let P(n) is true for n=k.

`P(k):3^(2k+2)-8k-9` is divisible by 8.

`3^(2k+2)-8k-9=8m` (say)

We wish to prove

P(k+1) is true whenever P(k) is true.

`P(k+1):3^(2(k+1)+2)-8(k+1)-9`

`=3^(2k+2+2)-8(k+1)-9`

`=3^2xx3^(2k+2)-8k-8-9`

`=9xx3^(2k+2)-72k+64k-9-8`

`=9(3^(2k+2)-8k-9)+64k+72-8`

`=9(8m)+64(k+1)`

and 72m and 64(k+1) are divisible by 8 because 72 and 64 are divisible by 8.

Thus P(k+1) is true when P(k) is true. Thus P(n) is true for all n, natural number.

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thanks