Prove that $2 \cdot 7^{n} + 3 \cdot 5^{n} - 5$ is divisible by $24$ for all $n \in N$.

(N/A) Let the statement $P(n)$ be defined as $P(n): 2 \cdot 7^{n} + 3 \cdot 5^{n} - 5$ is divisible by $24$.
Step $1$: For $n=1$,$2 \cdot 7^{1} + 3 \cdot 5^{1} - 5 = 14 + 15 - 5 = 24$,which is divisible by $24$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$,i.e.,$2 \cdot 7^{k} + 3 \cdot 5^{k} - 5 = 24q$ for some $q \in N$. This implies $2 \cdot 7^{k} = 24q - 3 \cdot 5^{k} + 5$.
Step $3$: We need to show $P(k+1)$ is true,i.e.,$2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5$ is divisible by $24$.
Consider $2 \cdot 7^{k+1} + 3 \cdot 5^{k+1} - 5 = 7(2 \cdot 7^{k}) + 15 \cdot 5^{k} - 5$.
Substitute $2 \cdot 7^{k} = 24q - 3 \cdot 5^{k} + 5$:
$= 7(24q - 3 \cdot 5^{k} + 5) + 15 \cdot 5^{k} - 5$
$= 168q - 21 \cdot 5^{k} + 35 + 15 \cdot 5^{k} - 5$
$= 168q - 6 \cdot 5^{k} + 30$
$= 168q - 6(5^{k} - 5)$.
Since $5^{k} - 5$ is divisible by $4$ for all $k \in N$ (as $5^{k} - 5 = 5(5^{k-1} - 1)$ and $5^{k-1} - 1$ is a multiple of $4$),let $5^{k} - 5 = 4m$.
Then the expression becomes $168q - 6(4m) = 168q - 24m = 24(7q - m)$,which is divisible by $24$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.