回復 2# shmilyho 的帖子
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\( \frac{1}{(4k-1)(4k+3)} = \frac{1}{4} \left( \frac{1}{4k-1} - \frac{1}{4k+3} \right) \)
故原式 \( = \frac{1}{4} \left( \frac{1}{3} - \frac17 + \frac17 -\frac1{11} + \ldots \frac{1}{4n-1} -\frac{1}{4n+3} \right) \)
\( = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{4n+3} \right) = \frac{n}{3(4n+3)} \)
