Q.Prove by method of induction, for all $n \in N$: $1^2 + 3^2 + 5^2 + \ldots + (2n-1)^2 = \dfrac{n}{3}(2n-1)(2n+1)$.

Let $P(n):1^2+3^2+\cdots+(2n-1)^2=\dfrac n3(2n-1)(2n+1)$. Base: $n=1$: L.H.S.$=1$, R.H.S.$=\dfrac13(1)(3)=1$; holds. Hypothesis: assume true for $k$. Step: add $(2k+1)^2$: $\dfrac k3(2k-1)(2k+1)+(2k+1)^2=(2k+1)\left[\dfrac{k(2k-1)}{3}+(2k+1)\right]=(2k+1)\cdot\dfrac{2k^2-k+6k+3}{3}=(2k+1)\cdot\dfrac{2k^2+5k+3}{3}=(2k+1)\cdot\dfrac{(k+1)(2k+3)}{3}=\dfrac{(k+1)}{3}(2k+1)(2k+3)$, which is $\dfra …

Identify $k$th odd number squared as $(2k-1)^2$, add the $(k+1) …

Confusing $(2n-1)^2$ (the term) with $(2n+1)$ (part of the formula); …