1/35+1/63+1/99+1/143 简便计算

1/35+1/63+1/99+1/143=1/(5×7)+1/(7×9)+1/(9×11)+1/(11×13)=1/2×(1/5-1/7)+1/2×(1/7-1/9)+1/2×(1/9-1/11)+1/2×(1/11-1/13)=1/2×(1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13)=1/2×(1/5-1/13)=1/2×(13/65-5/65)=1...为啥会有1/21/n-1/(n+2)
=(n+2)/[n(n+2)]-n/[n(n+2)]
=2/[n(n+2)]

∴1/[n(n+2)]=1/2×[1/n-1/(n+2)]