1/[n(n+1)(n+2)]
= [1/n - 1/(n+1)]*[1/(n+2)]
= [1/n - 1/(n+2)]/2 -[1/(n+1) - 1/(n+2)]
= [1/n + 1/(n+2)]/2 -[1/(n+1)]
原式
= (1/1 + 1/3)/2 - 1/2
+ (1/2 + 1/4)/2 - 1/3
+ (1/3 + 1/5)/2 - 1/4
+ ...
+ (1/18 + 1/20)/2 - 1/19
= ...
利用式子
1/[n(n+1)(n+2)]=(1/2){1/[n(n+1)]-1/[(n+1)(n+2)]}
就可以啦!