Question

裂项相消法：1/[n(n+1)(n+2)]=1/2*{1/[n(n+1)]-1/[(n+1)(n+2)}如何理解？急用，高手赐教！

裂项相消法：1/[n(n+1)(n+2)]=1/2*{1/[n(n+1)]-1/[(n+1)(n+2)}如何理解？急用，高手赐教！尽量详细。

Answer 1

简单计算一下即可，详情如图所示

Answer 2

1/[n(n+1)(n+2)]=(1/2)(n+2-n)/[n(n+1)(n+2)]=(1/2){(n+2)/[n(n+1)(n+2)-n/[n(n+1)(n+2)]}=(1/2){1/[n(n+1)]-1/[(n+1)(n+2)]}