## Precalculus (6th Edition) Blitzer

The required solution is $\left( k+1 \right)\left( 2k+3 \right)$.
Let us consider the statement $3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right)$ If $n=1$, the statement is $3=1\left( 2+1 \right)$ If $n=2$, the statement is $3+7=2\left( 4+1 \right)$ If $n=3$, the statement is $3+7+11=3\left( 6+1 \right)$. If $n=k+1$, the statement before the algebra is simplified is $3+7+11+\ldots +\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right)$. If $n=k+1$, the statement after the algebra is simplified is $3+7+11+\ldots +\left( 4k+3 \right)=\left( k+1 \right)\left( 2k+3 \right)$. For the statement $3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right)$, and if $n=1$ putting, \begin{align} & 3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right) \\ & 3=1\left( 2\times 1+1 \right) \\ & 3=1\left( 2+1 \right) \end{align} If the value of $n=2$ then we get, \begin{align} & 3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right) \\ & 3+\ldots +\left( 4\times 2-1 \right)=2\left( 2\times 2+1 \right) \\ & 3+\ldots +\left( 8-1 \right)=2\left( 4+1 \right) \\ & 3+7=2\left( 4+1 \right) \end{align} After that, putting $n=3$, we get \begin{align} & 3+7+11+\ldots +\left( 4n-1 \right)=n\left( 2n+1 \right) \\ & 3+\ldots +\left( 4\times 3-1 \right)=3\left( 2\times 3+1 \right) \\ & 3+\ldots +\left( 12-1 \right)=3\left( 6+1 \right) \\ & 3+7+11=3\left( 5+1 \right) \end{align} If $n=k+1$ substituting, $3+7+11+\ldots +\left( 4k-1 \right)=k\left( 2k+1 \right)$ Further simplification gives, \begin{align} & 3+7+11+\ldots +\left( 4\left( k+1 \right)-1 \right)=\left( k+1 \right)\left( 2\left( k+1 \right)+1 \right) \\ & 3+7+11+\ldots +\left( 4k+4-1 \right)=\left( k+1 \right)\left( 2k+2+1 \right) \\ & 3+7+11+\ldots +\left( 4k+3 \right)=\left( k+1 \right)\left( 2k+3 \right) \end{align}