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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $ 3^{2n+2}-8n-9$ is divisible by 8.

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Let the given statement be $P(n)$, i.e.,
$P(n) : 3^{2n+2}-8n-9$ is divisible by 8.
It can be observed that $P(n) $ is true for $n=1$ Since $ 3^{2 \times 1+2}-8 \times 1-9 = 64$, which is divisible by 8.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$3^{2k+2}-8k-9$ is divisible by 8.
$ \therefore 3^{2k+2}-8k-9 =8m$; where, $m \in N$------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
$= 3^{2k+2}.3^2-8k-8-9$
$ = 3^2 (3^{2k+2}-8k-9+8k+9)-8k-17$
$ = 3^2(3^{2k+2}-8k-9)+3^2(8k+9)-8k-17$
$ = 9.8m+9(8k+9)-8k-17$
$ = 9.8m+72k+81-8k-17$
$ = 9.8m+64k+64$
$ = 8(9m+8k+8)$
$ = 8r$ where $r=(9m+8k+8)$ is a natural number.
Therefore, $3^{2(k+1)+2}-8(k+1)-9$ is divisible by 8.
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered May 1, 2014 by thanvigandhi_1

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