32n+7 is divisible by 8 for all n ∈ N.
Let P(n) be the given statement.
Now,
$P(n): 3^{2 n}+7$ is divisible by 8 for all $n \in N$
Step 1:
$P(1)=3^{2}+7=9+7=16$
It is divisible by 8 .
Step 2:
Let $P(m)$ be true.
Then, $3^{2 m}+7$ is divisible by 8 . T
hus, $3^{2 m}+7=8 \lambda$ for some $\lambda \in N$. ....(1)
We need to show that $P(m+1)$ is true whenever $P(m)$ is true.
Now,
$P(m+1)=3^{2 m+2}+7$
$=3^{2 m} \cdot 9+7$
$=(8 \lambda-7) \cdot 9+7$ From (1)
$=72 \lambda-63+7$
$=72 \lambda-56$
$=8(9 \lambda-7)$
It is a multiple of $8 .$
Thus, $P(m+1)$ is divisible by 8 .
By the principle of $m$ athematical induction, $P(n)$ is true for all $n \in N$.
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