Q.Prove that $2.7^n + 3.5^n - 5$ is divisible by 24, for all $n \in N$.

Let $P(n)$ be the statement: $2\cdot7^n+3\cdot5^n-5$ is divisible by $24$.

Base case: For $n=1$,

divisible by $24$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$, i.e. there is an integer $m$ with

We must show $2\cdot7^{k+1}+3\cdot5^{k+1}-5$ is divisible by $24$.

Write $7^{k+1}=7\cdot7^k=7^k+6\cdot7^k$ and $5^{k+1}=5\cdot5^k=5^k+4\cdot5^k$. Then

Adding these and subtracting $5$:

$=24m+12\big(7^k+5^k\big)$ …

Divisibility induction with a parity trick: split $7^{k+1}$ and $5^{k+1}$ as (previous power) $+$ (multiple of the previous power) so the induction hypothesis appears directly, then show the leftover term …

• Trying to factor $2\cdot7^{k+1}+3\cdot5^{k+1}-5$ as $7\times(2\cdot7^k+3\cdot5^k-5)$ directly — this does not work because the coefficient of $5^k$ does not scale by $7$; the split into $6\cdot7^k$ and $4\cdot5^k$ is needed instead. …