1/n(n+1)=1/n-1/(n+1)
1/(2n-1)(2n+1)=1/2[1/(2n-1)-1/(2n+1)]
1/n(n+1)(n+2)=1/2[1/n(n+1)-1/(n+1)(n+2)]
1/(√a+√b)=[1/(a-b)](√a-√b)
n·n!=(n+1)!-n!
扩展资料:
【例1】【分数裂项基本型】求数列an=1/n(n+1) 的前n项和.
解:an=1/[n(n+1)]=(1/n)- [1/(n+1)](裂项)
则 Sn=1-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)…+(1/n)- [1/(n+1)](裂项求和)
= 1-1/(n+1)
= n/(n+1)
【例2】【整数裂项基本型】求数列an=n(n+1) 的前n项和.
解:an=n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3(裂项)
则 Sn=[1×2×3-0×1×2+2×3×4-1×2×3+……+n(n+1)(n+2)-(n-1)n(n+1)]/3(裂项求和)
= [n(n+1)(n+2)]/3