Solve the following


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


Let P(n) be the given statement.


$P(n): 3^{2 n}+7$ is divisible by 8 for all $n \in N$

Step 1:


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.


$P(m+1)=3^{2 m+2}+7$

$=3^{2 m} \cdot 9+7$

$=(8 \lambda-7) \cdot 9+7$    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 $n \in N$.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now