Q.Prove, by method of induction, for all $n \in N$: $\dfrac{1}{3.4.5} + \dfrac{2}{4.5.6} + \dfrac{3}{5.6.7} + \ldots + \dfrac{n}{(n+2)(n+3)(n+4)} = \dfrac{n(n+1)}{6(n+3)(n+4)}$.

Let $P(n):\dfrac1{3.4.5}+\dfrac2{4.5.6}+\cdots+\dfrac n{(n+2)(n+3)(n+4)}=\dfrac{n(n+1)}{6(n+3)(n+4)}$. Base: $n=1$: L.H.S.$=\dfrac1{60}$, R.H.S.$=\dfrac{2}{6(4)(5)}=\dfrac1{60}$; holds. Hypothesis: assume true for $k$. Step: add $\dfrac{k+1}{(k+3)(k+4)(k+5)}$: $\dfrac{k(k+1)}{6(k+3)(k+4)}+\dfrac{k+1}{(k+3)(k+4)(k+5)}=\dfrac{k+1}{(k+3)(k+4)}\left[\dfrac k6+\dfrac1{k+5}\right]=\dfrac{k+1}{(k+3)(k+4)}\cdot\dfrac{k(k+5)+6}{6(k+5)}$. Since $k(k+5)+6=k^2+5k+6=(k+2)(k+3)$, this becomes $\dfrac{(k+1)(k+2)(k+3)}{6(k+3)(k+4)(k+5)}=\d …

Add the $(k+1)$th term, combine over a common denominator, factor the numerator $k^2+5k …

Not spotting the key factorisation $(k+2)(k+3)$ that lets the $(k+3 …