If 3^{2n+2}-8n-9 is divisible by 2^p for all | vedclass

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(C) Let $f(n) = 3^{2n+2}-8n-9 = 9(3^{2n})-8n-9 = 9(9^n)-8n-9$. Using the Binomial Theorem,$9^n = (1+8)^n = 1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots +

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let $f(n) = 3^{2n+2}-8n-9 = 9(3^{2n})-8n-9 = 9(9^n)-8n-9$. Using the Binomial Theorem,$9^n = (1+8)^n = 1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots + 8^n$. Substituting this into the expression: $f(n) = 9[1 + 8n + 64 \cdot \frac{n(n-1)}{2} + \dots] - 8n - 9$ $f(n) = 9 + 72n + 9 \cdot 64 \cdot \frac{n(n-1)}{2} + \dots - 8n - 9$ $f(n) = 64n + 288n(n-1) + \dots$ $f(n) = 64n + 288(n^2-n) + \dots$ Since $64 = 2^6$ and $288 = 32 	imes 9 = 2^5 	imes 9$,the expression is divisible by $2^6 = 64$. For $n=1$,$f(1) = 3^4 - 8(1) - 9 = 81 - 17 = 64 = 2^6$. For $n=2$,$f(2) = 3^6 - 8(2) - 9 = 729 - 16 - 9 = 704 = 64 	imes 11$. Thus,the maximum value of $p$ is $6$.

If $3^{2n+2}-8n-9$ is divisible by $2^p$ for all $n \in N$,then the maximum value of $p$ is

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