Q.Prove by method of induction, for all $n \in N$: $1.3 + 3.5 + 5.7 + \ldots \text{ to } n \text{ terms} = \dfrac{n}{3}(4n^2+6n-1)$.

Let $P(n):1.3+3.5+\cdots\text{ to }n\text{ terms}=\dfrac n3(4n^2+6n-1)$, $k$th term $(2k-1)(2k+1)=4k^2-1$. Base: $n=1$: L.H.S.$=3$, R.H.S.$=\dfrac13(4+6-1)=3$; holds. Hypothesis: assume true for $k$. Step: add $4(k+1)^2-1$: $\dfrac k3(4k^2+6k-1)+4(k+1)^2-1=\dfrac{4k^3+6k^2-k+12k^2+24k+11}{3}=\dfrac{4k^3+18k^2+23k+11}{3}$. The target at $n=k+1$: $\dfrac{k+1}{3}(4(k+1)^2+6(k+1)-1)=\dfrac{k+1}{3}(4k^2+14k+9)=\dfrac{4k^3+14k^2+9k+4k^2+14k+9}{3}=\dfrac{4k^3+18k^2+23k+9}{3}$; recomp …

Recognise the $k$th term as $4k^2-1$, add to the assumed sum, expand and match again …

Sign errors expanding $4(k+1)^2-1$; losing track of terms when combining …