求和:1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+...+1/[(2n)^2-1]
问题描述:
求和:1/(2^2-1)+1/(4^2-1)+1/(6^2-1)+...+1/[(2n)^2-1]
答
很简单
每项都可以分解的
1/(2^2-1)=1/2(1-1/2)
1/(4^2-1)=1/2(1/3-1/5)
总的就=1/2(1-1/2+1/3-1/5+1/5-1/7+1/7-1/9+...-1/(2n+1))
整理下=5/12+1/(4n+2)