对于任意的正整数n,有1/1*2*3 + 1/2*3*4 +...1/n(n+1)(n+2)

1/n(n+1)(n+2)
=[1/n(n+1)-1/(n+1)(n+2)]/2
1/1*2*3 + 1/2*3*4 +...+1/n(n+1)(n+2)
=[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]/2
=[1/2-1/(n+1)(n+2)]/2
=1/4-1/2(n+1)(n+2)对于任意的正整数n，有1/1*2*3 + 1/2*3*4 +...+1/n(n+1)(n+2)

1/n(n+1)(n+2) =1/2[1/n（n+1）-1/（n+1)(n+2）]
1/1*2*3 + 1/2*3*4 +...1/n(n+1)(n+2)
=1/2[1/1*2-1/2*3+1/2*3-1/3*4+.....1/n（n+1）-1/（n+1)(n+2）]
=1/2[1/1*2-1/（n+1)(n+2）]
=0.25-1/2*1/（n+1)(n+2）

解 1/n(n+1)(n+2)=[1/n-1/n+1](1/n+1)=1/n(n+1)-1/(n+1)(n+2)=1/2(1/n-1/n+2)-1/n+1+1/n+2=1/2[1/n+1/n+2-2/n+1]=1/2[1/n-1/n+1+1/n+2-1/n+1]所以 1/1*2*3 +1/2*3*4+...+1/n(n+1)(n+2)=1/2(1-1/2+1/3-1/2+1/2-1/3+1/...