Answer
$\therefore \sum_{k=2}^{20}\left(-\frac{1}{(k-1)^{2}}+\frac{1}{k^{2}}\right)=-\frac{399}{400}$
Work Step by Step
We find:
$\begin{aligned} \sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right) &=\left(\frac{1}{2^{2}}-\frac{1}{(2-1)^{2}}\right)+\left(\frac{1}{3^{2}}-\frac{1}{(3-1)^{2}}\right)+\left(\frac{1}{4^{2}}-\frac{1}{(4-1)^{2}}\right)+\ldots+\left(\frac{1}{20^{2}}-\frac{1}{(20-1)^{2}}\right) \\ &=\left(\frac{1}{2^{2}}-\frac{1}{1^{2}}\right)+\left(\frac{1}{3^{2}}-\frac{1}{2^{2}}\right)+\left(\frac{1}{4^{2}}-\frac{1}{3^{2}}\right)+\ldots+\left(\frac{1}{20^{2}}-\frac{1}{19^{2}}\right) \\ &=\left(\frac{1}{4}-\frac{1}{1}\right)+\left(\frac{1}{9}-\frac{1}{4}\right)+\left(\frac{1}{16}-\frac{1}{9}\right)+\ldots+\left(\frac{1}{400}-\frac{1}{361}\right) \end{aligned}$
$\left(\frac{y}{1}-\frac{1}{1}\right)+\left(\frac{y}{9}-\frac{y}{1}\right)+\left(\frac{1}{16}-\frac{y}{9}\right)+\ldots+\left(-\frac{1}{361}+\frac{1}{100}\right)$
$=-1+\frac{1}{400}$
$=-\frac{399}{400}$
$\therefore \sum_{k=2}^{20}\left(\frac{1}{k^{2}}-\frac{1}{(k-1)^{2}}\right)=-\frac{399}{400}$
