Q.By principle of mathematical induction prove that $1^3+2^3+3^3+\ldots\ldots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$ OR For every positive integer $n$, prove that $7^n - 3^n$ is divisible by 4.

By the principle of mathematical induction, $1^3+2^3+\cdots+n^3=\left[\dfrac{n(n+1)}{2}\right]^2$ holds for every natural number $n$.

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

LHS $=$ RHS, so $P(1)$ is true.

Step 2: Inductive step

Assume $P(k)$ is true for some $k\in\mathbb{N}$, i.e.

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

We must show $P(k+1)$ is true:

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

Starting from the LHS and using the assumption:

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

Factor out $(k+1)^2$:

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

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

This is exactly the RHS of $P(k+1)$. So $P(k+1)$ is true whenever $P(k)$ is true.

Step 3: Conclusion …