## Precalculus (6th Edition) Blitzer

The required solution is ${{S}_{k}}=k\left( 2k+1 \right),{{S}_{k+1}}=(k+1)(2k+3)$
Then using ${{S}_{k}},{{S}_{k+1}}$ from the expression ${{S}_{n}}:3+7+11+...+\left( 4n-1 \right)=n\left( 2n+1 \right)$ ${{S}_{k}}$ is provided by, $3+7+11+...+\left( 4k-1 \right)=k\left( 2k+1 \right)$ ${{S}_{k+1}}$ is given by, \begin{align} & 3+7+11+...+\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right) \\ & =\left( k+1 \right)\left( 2k+2+1 \right) \\ & =\left( k+1 \right)\left( 2k+3 \right) \\ \end{align} We get the solution ${{S}_{k}}=k\left( 2k+1 \right),{{S}_{k+1}}=(k+1)(2k+3)$.