$\begin{align} & \text{Find the }{{r}^{th}}\text{ term of the series }(3\times 4)+(4\times 5)+(5\times 6)+\cdot \cdot \cdot \\ & \text{Hence, find the sum of the first }n\text{ terms of the series} \\ & \text{Verify your answers by setting }n=3 \\\end{align}$
$\begin{align} & \text{ }(3\times 4)+(4\times 5)+(5\times 6)+\cdot \cdot \cdot (r+2)(r+3) \\ & {{T}_{r}}=(r+2)(r+3)={{r}^{2}}+5r+6n \\ & {{T}_{n}}={{n}^{2}}+5n+6 \\ & {{S}_{n}}=\sum{{{T}_{n}}}=\sum {{n}^{2}}+5\sum n+6n \\ & {{S}_{n}}=\tfrac{n}{6}(n+1)(2n+1)+\frac{5n(n+1)}{2}+6n \\ & {{S}_{n}}=n\left[ \frac{(n+1)(2n+1)}{6}+\frac{5(n+1)}{2}+6 \right] \\ & {{S}_{n}}=n\left[ \frac{(n+1)}{2}\left( \frac{2n+1}{3}+5 \right)+6 \right] \\ & \text{When }n=3 \\ & {{S}_{3}}=3\left[ \frac{(3+1)}{2}\left( \frac{2(3)+1}{3}+5 \right)+6 \right] \\ & {{S}_{3}}=3\left[ 2\left( \frac{7}{3}+5 \right)+6 \right]=62 \\ & \text{Also } \\ & {{S}_{3}}=(3\times 4)+(4\times 5)+(5\times 6)=12+20+30=62 \\\end{align}$