For every natural number n,{3^{2n + 2}} - 8n | vedclass

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Question

(A) Let $P(n) = 3^{2n+2} - 8n - 9$. For $n=1$,$P(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64$,which is divisible by $16$. For $n=2$,$P(2) = 3^6 - 8(2) - 9 = 72

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let $P(n) = 3^{2n+2} - 8n - 9$. For $n=1$,$P(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64$,which is divisible by $16$. For $n=2$,$P(2) = 3^6 - 8(2) - 9 = 729 - 16 - 9 = 704$. $704 = 16 	imes 44$,so it is divisible by $16$. Using the Binomial Theorem: $3^{2n+2} = 9 	imes 3^{2n} = 9 	imes (1+8)^n$. $9 	imes (1+8)^n = 9 	imes [1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots]$. $9 	imes [1 + 8n + 32n(n-1) + \dots] = 9 + 72n + 288n(n-1) + \dots$. Thus,$P(n) = 9 + 72n + 288n(n-1) + \dots - 8n - 9 = 64n + 288n(n-1) + \dots$. Since every term is divisible by $64$,the expression is divisible by $64$,and consequently by $16$.

For every natural number $n$,${3^{2n + 2}} - 8n - 9$ is divisible by

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