(1)1/[n(n+1)]=(1/n)- [1/(n+1)](2)1/[(2n-1)(2n+1)]=1/2[1/(2n-1)-1/(2n+1)](3)1/[n(n+1)(n+2)]=1/2{1/[n(n+1)]-1/[(n+1)(n+2)]}(4)1/(√a+√b)=[1/(a-b)](√a-√b)(5) n·n!=(n+1)!-n!(6)1/[n(n+k)]=1/k[1/n-1/(n+k)](7)1/[√n+√(n+1)]=√(n+1)-√n(8)1/(√n+√n+k)=(1/k)·[√(n+k)-√n]【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【例3】1/(1×4)+1/(4×7)+1/(7×10)+……+1/(91×94)