nx(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]
1x2+2x3+3x4+...+nx(n+1)=1/3[1x2x3-0x1x2+2x3x4-1x2x3+3x4x5- 2x3x4+...+n(n+1)(n+2)-(n-1)n(n+1)]=1/3[n(n+1)(n+2)-0x1x2]