裂项相消法常见公式1/n(n+1)(n+2)=?
问题描述:
裂项相消法常见公式
1/n(n+1)(n+2)=?
答
1/n(n+1)(n+2)=0.5[1/n(n+1)-1/(n+1)(n+2)]
答
1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)]
这样才能多项 相消
答
原式=1/[n(n+1)]-1/[n(n+2)]
=1/n-1/(n+1)-[1/2n-1/2(n+2)]
=1/2n-1/(n+1)+1/2(n+2)
答
1