For all positive integral values of n,{3^{2n} | vedclass

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Question

(A) Let $f(n) = 3^{2n} - 2n + 1$. For $n = 1$,$f(1) = 3^{2(1)} - 2(1) + 1 = 9 - 2 + 1 = 8$. For $n = 2$,$f(2) = 3^{2(2)} - 2(2) + 1 = 81 - 4 + 1 = 78$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $f(n) = 3^{2n} - 2n + 1$. For $n = 1$,$f(1) = 3^{2(1)} - 2(1) + 1 = 9 - 2 + 1 = 8$. For $n = 2$,$f(2) = 3^{2(2)} - 2(2) + 1 = 81 - 4 + 1 = 78$. Wait,let us re-evaluate the expression. $3^{2n} = 9^n = (1 + 8)^n = 1 + n(8) + \frac{n(n-1)}{2}(8^2) + \dots$ So,$3^{2n} - 2n + 1 = (1 + 8n + 32n(n-1) + \dots) - 2n + 1 = 6n + 2 + 32n(n-1) + \dots$ Actually,checking $n=1$ gives $8$,which is divisible by $8$. Checking $n=2$ gives $78$,which is not divisible by $8$. However,the expression $3^{2n} - 2n + 1$ is always divisible by $8$ if the term was $3^{2n} - 8n + 1$ or similar. Given the options,let us check $n=1 \implies 8$ and $n=2 \implies 78$. Since $8$ is divisible by $2$ and $78$ is divisible by $2$,the correct option is $2$.

For all positive integral values of $n$,${3^{2n}} - 2n + 1$ is divisible by

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