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数学公式再次测试

n=0(n!)22n+1(2n+1)!=n=010tn(1?t)n2n+1dt=2100tn(1?t)n2ndt=101(t?12)+14dt=2 arctan2(t?12)|10=π
<script id="MathJax-Element-1" type="math/tex; mode=display">\begin{align*}\sum_{n=0}^{\infty}\frac{(n!)^{2}2^{n+1}}{(2n+1)!}&=\sum_{n=0}^{\infty}\int_{0}^{1}t^{n}(1-t)^{n}2^{n+1}dt\\&=2\int_{0}^{1}\sum_{0}^{\infty}t^{n}(1-t)^{n}2^{n}dt\\&=\int_{0}^{1}\frac{1}{(t-\frac{1}{2})+\frac{1}{4}}dt\\&=2 \arctan2(t-\frac{1}{2})|_{0}^{1}\\&=\pi\end{align*}</script>

 

特为http://www.cnblogs.com/zhangwenbiao/p/3705281.html写的代码.