回復 1# ycdye 的帖子
第12題
\(\begin{align}
& \sum\limits_{k=1}^{1000}{\frac{{{k}^{2}}}{\left( 2k-1 \right)\left( 2k+1 \right)}} \\
& =\sum\limits_{k=1}^{1000}{\left( \frac{1}{4}+\frac{\frac{1}{4}}{4{{k}^{2}}-1} \right)} \\
& =\sum\limits_{k=1}^{1000}{\left[ \frac{1}{4}+\frac{1}{8}\left( \frac{1}{2k-1}-\frac{1}{2k+1} \right) \right]} \\
& =250+\frac{1}{8}\left( 1-\frac{1}{2001} \right) \\
& =250\frac{250}{2001} \\
\end{align}\)
