求和 sn＝1×2×3+2×3×4+……+n(n+1)(n+2)
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如果是简单数列问题可以这样做n(n+1)(n+2)=n^3+3n^2+2n
1^3+2^3+……+n^3=[n(n+1)/2]^2=n^2(n+1)^2/4
1^2+2^2+……+n^2=n(n+1)(2n+1)/6
1+2+……+n=n(n+1)/2
所以1*2*3+2*3*4+……+n(n+1)(n+2)
=n^2(n+1)^2/4+3*n(n+1)(2n+1)/6+2*n(n+1)/2
=n^2(n+1)^2/4+n(n+1)(2n+1)/2+n(n+1)
=[n(n+1)/4][n(n+1)+2(2n+1)+4]
=[n(n+1)/4](n^2+n+4n+2+4)
=[n(n+1)/4](n^2+5n+6)
=n(n+1)(n+2)(n+3)/4
1^3+2^3+……+n^3=[n(n+1)/2]^2=n^2(n+1)^2/4
1^2+2^2+……+n^2=n(n+1)(2n+1)/6
1+2+……+n=n(n+1)/2
所以1*2*3+2*3*4+……+n(n+1)(n+2)
=n^2(n+1)^2/4+3*n(n+1)(2n+1)/6+2*n(n+1)/2
=n^2(n+1)^2/4+n(n+1)(2n+1)/2+n(n+1)
=[n(n+1)/4][n(n+1)+2(2n+1)+4]
=[n(n+1)/4](n^2+n+4n+2+4)
=[n(n+1)/4](n^2+5n+6)
=n(n+1)(n+2)(n+3)/4
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