7楼: Originally posted by Edstrayer at 2014-10-26 13:02:45
设
a_n=\cos^{2n}\left(\frac{\pi}{8}\right)+\sin^{2n}\left(\frac{\pi}{8}\right)(n=1,2,\cdots)
则有：
a_n=\frac{1}{4^n}\left
所以就有：
a_n=\frac{\sum\limits_{i=0}^{\left}{n \choose 2i}2^{n+1-i}}{ ...