Solve the following
Question:

32n+7 is divisible by 8 for all n ∈ N.

Solution:

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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