写一个一般的等式n×(n+1)=1/3[n(n+1)(n+2)-(n-1)n(n+1)]对吧两边约去n(n+1)，实际上是1=1/3[(n+2)-(n-1)]所以写成三项的话1=1/4[(n+3)-(n-1)]n(n+1)(n+2)=1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]叠加就可以了
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