2.7n + 3.5n − 5 is divisible by 24 for all n ∈ N.
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 $n \in N$.
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