## Precalculus (6th Edition) Blitzer

The required solution is ${{S}_{k}}=\frac{k\left( 5k-1 \right)}{2},{{S}_{k+1}}=\frac{\left( 5{{k}^{2}}+9k+4 \right)}{2}$
Now, using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:2+7+12+...+\left( 5n-3 \right)=\frac{n\left( 5n-1 \right)}{2}$ ${{S}_{k}}$ is provided by, $2+7+12+...+\left( 5k-3 \right)=\frac{k\left( 5k-1 \right)}{2}$ ${{S}_{k+1}}$ is provided by, \begin{align} & 2+7+12+...+\left( 5\left( k+1 \right)-3 \right)=\frac{\left( k+1 \right)\left( 5\left( k+1 \right)-1 \right)}{2} \\ & =\frac{\left( k+1 \right)\left( 5k+5-1 \right)}{2} \\ & =\frac{\left( k+1 \right)\left( 5k+4 \right)}{2} \\ & =\frac{\left( 5{{k}^{2}}+9k+4 \right)}{2} \end{align} Then, we get the solution ${{S}_{k}}=\frac{k\left( 5k-1 \right)}{2},{{S}_{k+1}}=\frac{\left( 5{{k}^{2}}+9k+4 \right)}{2}$.