Q.Prove the following by using the principle of Mathematical induction for all $n \in N$: $1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$

The formula holds for $n=1$, and assuming it for $n=k$ forces it for $n=k+1$ — so it holds for all $n \in N$ by the principle of mathematical induction.

Let $P(n): 1^2+3^2+5^2+\cdots+(2n-1)^2 = \dfrac{n(2n-1)(2n+1)}{3}$.

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

Inductive step: Assume $P(k)$ is true:

$1^2+3^2+\cdots+(2k-1)^2 = \frac{k(2k-1)(2k+1)}{3}$

Add the next term $(2k+1)^2$ (which is $(2(k+1)-1)^2$) to both sides:

$1^2+3^2+\cdots+(2k-1)^2+(2k+1)^2 = \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2$

$= (2k+1)\left[\frac{k(2k-1)}{3}+(2k+1)\right] = (2k+1)\cdot\frac{k(2k-1)+3(2k+1)}{3}$

$= (2k+1)\cdot\frac{2k^2-k+6k+3}{3} = (2k+1)\cdot\frac{2k^2+5k+3}{3}$

Factoring $2k^2+5k+3=(2k+3)(k+1)$: …