如何对1除以n(n+1)(n+2)进行裂项求和?

如何对1除以n(n+1)(n+2)进行裂项求和?
如何对1除以n(n+1)(n+2)进行裂项求和?

如何对1除以n(n+1)(n+2)进行裂项求和?
通常用待定系数法,可设1/n*(n+1)*(n+2)=a/n+b/(n+1)+c/(n+2)
两边去掉分母后对比各项系数,解出系数即可.
1/n*(n+1)*(n+2)=0.5/n-1/(n+1)+0.5/(n+2)
Sn=[1-1/2-1/(n+1)+1/(n+2)]/2=[1/2-1/(n+1)+1/(n+2)]/2

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