1/1*2*3+1/2*3*4+1/3*4*5+……+1/11*12*13

1/n(n+1)(n+2) = [1/n(n+1) - 1/(n+1)(n+2)]/2
按这个思路提示就简单了。
结果应该是 (1/2 - 1/12*13)/2 = 1/4-1/(13*24) = 77/312

1/n(n+1)(n+2) = [1/n(n+1) - 1/(n+1)(n+2)]/2
原式=(1/1*2-1/2*3+1/2*3-1/3*4+......+1/11*12-1/12*13)/2
=(1/2-1/12*13)/2
=77/312

1/n(n+1)(n+2) = [1/n(n+1) - 1/(n+1)(n+2)]/2
=[(1/n-1/(n+1))-(1/(n+1)-1/(n+2))]/2
裂项相消，
原式=1/2(1-1/2-1/(11+1)+1/(11+2))
=(1/2)*(77/156)=77/312