Q.(a) Prove that for any natural number $n$, $a^n-b^n$ is divisible by $a-b$, where $a>b$. OR

The base case $n=1$ is immediate; the inductive step rewrites $a^{k+1}-b^{k+1}$ as $a(a^k-b^k)+b^k(a-b)$, both terms of which are divisible by $(a-b)$, completing the induction.

Statement: For every natural number $n$, $a^n-b^n$ is divisible by $(a-b)$, where $a>b$.

Base case ($n=1$): $a^1-b^1 = a-b$, which is divisible by $(a-b)$ (quotient 1). True.

Inductive hypothesis: Assume the statement is true for $n=k$, i.e. $a^k-b^k = (a-b)\cdot m$ for some integer $m$ (in other words, $a^k-b^k$ is divisible by $a-b$).

Inductive step ($n=k+1$): We want to show $a^{k+1}-b^{k+1}$ is divisible by $(a-b)$.

$a^{k+1}-b^{k+1} = a\cdot a^k - b\cdot b^k = a\cdot a^k - a\cdot b^k + a\cdot b^k - b\cdot b^k$

$= a(a^k-b^k) + b^k(a-b)$

By the inductive hypothesis, $a^k-b^k=(a-b)m$, so:

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