## Precalculus (6th Edition) Blitzer

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; $\frac{1}{2\cdot 3}=\frac{1}{6}$ And simplifying on the right, we obtain $\frac{n}{2n+4}=\frac{1}{6}$. So, this true statement shows that ${{S}_{1}}$ is true. Also, assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, $\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}=\frac{k}{2k+4}$ By adding $\frac{1}{\left( k+2 \right)\left( k+3 \right)}$ on both sides as: \begin{align} & \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{k}{2k+4}+\frac{1}{\left( k+2 \right)\left( k+3 \right)} \\ & \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{k\left( k+3 \right)}{2\left( k+2 \right)\left( k+3 \right)}+\frac{2}{2\left( k+2 \right)\left( k+3 \right)} \\ & \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{{{k}^{2}}+3k+2}{2\left( k+2 \right)\left( k+3 \right)} \\ \end{align} Therefore, \begin{align} & \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{\left( k+1 \right)}{2\left( k+3 \right)} \\ & \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( k+1 \right)\left( k+2 \right)}+\frac{1}{\left( k+2 \right)\left( k+3 \right)}=\frac{\left( k+1 \right)}{2\left( k+1 \right)+4} \\ \end{align} Thus, ${{S}_{k+1}}$ is true. The result $\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{\left( n+1 \right)\left( n+2 \right)}=\frac{n}{2n+4}$ holds true.