Solve the following

Let P(n) be the given statement.

Now,

$P(n): 2.7^{n}+3.5^{n}-5$ is divisible by 24 .

Step 1:

$P(1): 2.7^{1}+3.5^{1}-5=24$

It is divisible by 24 .

Thus, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

Then,

2. $7^{m}+3.5^{m}-5$ is divisible by 24 .

Suppose :

$2.7^{m}+3.5^{m}-5=24 \lambda \quad \ldots(1)$

We need to show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1)=2.7^{m+1}+3.5^{m+1}-5$

$=2.7^{m+1}+\left(24 \lambda+5-2.7^{m}\right) 5-5$

$=\left(2.7^{m+1}+120 \lambda+25-10.7^{m}-5\right)$

$=2.7^{m} \cdot 7-10.7^{m}+120 \lambda+24-4$

$=7^{m}(14-10)+120 \lambda+24-4$

$=7^{m} \cdot 4+120 \lambda+24-4$

$=4\left(7^{m}-1\right)+24(5 \lambda+1)$

$=4 \times 6 \mu+24(5 \lambda+1)$

[Since $7^{m}-1$ is a multiple of 6 for all $n \in N, 7^{m}-1=\mu .$ ]

$=24(\mu+5 \lambda+1)$

It is a multiple of 24 .

Thus, $P(m+1)$ is true.

By the $p$ rinciple of $m$ athematical $i$ nduction, $P(n)$ is true for $n \in N$.