Q.(a) Prove by mathematical induction that $7^{2n} + 16n - 1$ is divisible by 64 for all $n \in N$. OR

By induction: the statement holds at $n=1$, and assuming it holds for $n=m$ forces it to hold for $n=m+1$, so it holds for every $n\in N$.

Let $P(n): 7^{2n}+16n-1$ is divisible by $64$.

Step 1 — Base case ($n=1$).

$7^{2(1)}+16(1)-1 = 49+16-1 = 64 = 64\times 1$, which is divisible by $64$. So $P(1)$ is true.

Step 2 — Inductive hypothesis.

Assume $P(m)$ is true for some $m\in N$, i.e. $7^{2m}+16m-1=64k$ for some integer $k$, so $7^{2m}=64k-16m+1$.

Step 3 — Inductive step, show $P(m+1)$.

$7^{2(m+1)}+16(m+1)-1 = 7^{2m+2}+16m+15 = 49\cdot7^{2m}+16m+15$

Substitute $7^{2m}=64k-16m+1$:

$=49(64k-16m+1)+16m+15$

$=49\times 64k - 784m + 49 + 16m + 15$

$=49\times 64k - 768m + 64$

$=64\big(49k-12m+1\big)$

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