Question

3²n + 7 is divisible by 8​

Answer 1

We're given to prove by principle of mathematical induction that,

$8\mid 3^{2n}+7,\quad n\in\mathbb{N}$

Let n = 1.

$3^2+7=9+7=16,\quad 8\mid 16$

So P(1) holds true. Assume so does P(k).

$\text{Let}\ \ 3^{2k}+7=8m,\quad m\in\mathbb{N}\\\\\implies\ \ 3^{2k}=8m-7\quad\quad(1)$

Consider P(k + 1).

$\begin{aligned}&3^{2(k+1)}+7\\\\\implies\ \ &3^{2k+2}+7\\\\\implies\ \ &3^{2k}\times 3^2+7\\\\\implies\ \ &(8m-7)9+7\quad[\text{From (1)}]\\\\\implies\ \ &8m\times 9-7\times 9+7\\\\\implies\ \ &8m\times 9+7(-9+1)\\\\\implies\ \ &8m\times 9-7\times 8\\\\\implies\ \ &8(9m-7)\\\\\implies\ \ &8\mid 3^{2(k+1)}+7\end{aligned}$

Hence Proved!

Answer 2

Step-by-step explanation:

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