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]
1x2=1/3(1x2x3=0x1x2 ) 2x3=1/3(2x3x4-1x2x3) 3x4=1/3(3x4x5- 2x
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