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

Base case n=1 holds; assuming the formula for n=k, adding (k+1)³ and factoring gives exactly the formula for n=k+1 — so by PMI it holds for every n≥1.

Let $P(n)$ be the statement

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

Step 1 — Base case ($n=1$):

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

Step 2 — Inductive hypothesis:

Assume $P(k)$ is true for some $k\geq1$, i.e.

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

Step 3 — Inductive step, show $P(k+1)$:

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

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

Factor out $(k+1)^2$ from the right side:

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

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

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