1/n(n+1)(n+2)(n+3)=1/[n(n+3)][(n+1)(n+2)]
=1/2[1/(n(n+3)--1/(n+1)(n+2)]
=1/2{1/3[(1/n--1/(n+3)]--[1/(n+1)--1/(n+2)]
=1/6(1/n)--1/6[1/(n+3)]--1/2[1/(n+1)]+1/2[1/(n+3)]
=1/6n--1/6(n--3)--1/2(n+1)+1/2(n+3).