求和:Sn=1/1×2×3+1/2×3×4+1/3×4×5+…+1/n×(n+1)×(n+2)

1/[n(n+1)(n+2)]
=1/2*{1/[n(n+1)] - 1/[(n+1)(n+2)]}
=1/2*[1/n -1/(n+1)] - 1/4*[1/(n+1)-1/(n+2)]
例如 1/2*3*4= 1/4*(1/2-1/3)-1/4*(1/3-1/4)
所以Sn= 1/2*[1/1-1/2+1/2-1/3+...+1/n-1/(n+1)] - 1/2*[1/2-1/3+1/3-1/4+...+1/(n+1)-1/(n+2)]
= 1/2*[1-1/(n+1)] - 1/2*[1/2-1/(n+2)]
= 1/2*[1/2 - 1/(n+1)(n+2)]

Sn=0.5[1/1×2-1/2×3+1/2×3-1/3×4+…+1/n×(n+1)-1/(n+1)×(n+2)]
=0.5[1/1×2-1/(n+1)×(n+2)]
=1/4-1/2(n+1)+1/2(n+2)

数列通项公式为an=1/n×(n+1)×(n+2) 时,一般用裂项法,就是把每一项分成两个项的差的形式.an=[1/n(n+1) - 1/(n+1)(n+2)]/2 ; （举例：an=1/n(n+1)=1/n - 1/(n+1); an=1/n(n+1)(n+2)(n+3)=[1/n(n+1)(n+2) - 1/(n+1)(n...

1/n×(n+1)×(n+2) = (1/n(n+2) - 1/(n+1)(n+2))= (1/2n - 1/2(n+2) -1/(n+1)+1/(n+2)) = 1/2n - 1/2(n+1) - 1/2(n+1) + 1/2(n+2)
Sn = Pn - Pn+1 + P1
其中Pn 是数列1/2n - 1/2(n+1)的前n项和
显然，Pn = 1/2 - 1/2(n+1)
Sn = Pn - Pn+1 = 1/2 - 1/2(n+1) - 1/2(1+1) + 1/2(n+2) = 1/4 - 1/2(n+1) + 1/2(n+2)