1*2+2*3.+9*10= 1*2+2*3.2008*2009= 1*2+2*3+.+n(n+1)= 1*2*3+2*
1个回答

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

1*2+2*3+3*4+.+n*(n+1)=1/3*1*2*3-0+1/3*2*3*4-1/3*1*2*3+1/3*3*4*5-1/3*2*3*4+.

1/3n(n+1)(n+2)-1/3(n-1)n(n+1)

=1/3*n(n+1)(n+2)

当n=9时,就是第一个和答案,当n等于2008时,就是第二个答案

几个连续数乘积之和的问题都能这样做

最后一个:

n(n+1)(n+2)=1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]

1*2*3+2*3*4.+n(n+1)(n+2)= 1/4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]

还可以进一步拓展到四个数相乘,或更多.

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