Question

Prove that 3^2n+2-8n-9 is divisible by 8 for all n>1.​
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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

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