Precalculus (6th Edition) Blitzer

The required value is ${{S}_{k}}=\frac{k\left( k+5 \right)}{2},{{S}_{k+1}}=\frac{\left( {{k}^{2}}+7k+6 \right)}{2}$
We are using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:3+4+5+...+\left( n+2 \right)=\frac{n\left( n+5 \right)}{2}$ ${{S}_{k}}$ is provided by, $3+4+5+...+\left( k+2 \right)=\frac{k\left( k+5 \right)}{2}$ ${{S}_{k+1}}$ is provided by, \begin{align} & 3+4+5+...+\left( k+1+2 \right)=\frac{\left( k+1 \right)\left( k+1+5 \right)}{2} \\ & =\frac{\left( k+1 \right)\left( k+6 \right)}{2} \\ & =\frac{\left( {{k}^{2}}+7k+6 \right)}{2} \end{align} Hence, the solution is ${{S}_{k}}=\frac{k\left( k+5 \right)}{2},{{S}_{k+1}}=\frac{\left( {{k}^{2}}+7k+6 \right)}{2}$.