Q.(a) By the principle of mathematical induction, prove that, for $n \ge 1$, $1^3+2^3+3^3+....+n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$. OR

Base case $n=1$ holds, and assuming the formula for $n=k$ lets us derive it for $n=k+1$, so by induction it holds for all $n\ge1$.

Let $P(n)$: $1^3+2^3+\cdots+n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$.

Base case ($n=1$): LHS $=1^3=1$. RHS $=\left(\dfrac{1\cdot2}{2}\right)^2=1^2=1$. So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true, i.e. $1^3+2^3+\cdots+k^3=\left(\dfrac{k(k+1)}{2}\right)^2$.

Add $(k+1)^3$ to both sides:

$1^3+\cdots+k^3+(k+1)^3 = \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3$

$= (k+1)^2\left[\dfrac{k^2}{4}+(k+1)\right] = (k+1)^2\cdot\dfrac{k^2+4k+4}{4} = (k+1)^2\cdot\dfrac{(k+2)^2}{4}$

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