如何证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3顺便再证明一下1x2+2x3+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

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如何证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3顺便再证明一下1x2+2x3+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4
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如何证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3顺便再证明一下1x2+2x3+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4
如何证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3
顺便再证明一下1x2+2x3+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

如何证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3顺便再证明一下1x2+2x3+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4
证明1x2+2x3+…+n(n+1)=n(n+1)(n+2)/3
1x2+2x3+…+n(n+1)=1x(1+1)+2x(2+1)+.+n(n+1)
=(1^2+2^2+.+n^2)+(1+2+.+n)
=n(n+1)(2n+1)/6 + n(n+1)/2
=n(n+1)(n+2)/3
证明1x2+2x3+…+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4是错的,我想应该是证明1x2x3+2x3x4+.+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4
若是证明1x2x3+2x3x4+.+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4
因为n(n+1)(n+2)=n^3+3n^2+2n
所以1x2x3+2x3x4+.+n(n+1)(n+2)
=(1^3+2^3+...+n^3) + 3(1^2+2^2+...+n^2) + 2(1+2+...+n)
=n^2(n+1)^2/4 + n(n+1)(2n+1)/2 + n(n+1)
=n(n+1)(n+2)(n+3)/4

用排列组合
左边=2[C(2)2+C(3)2+....+C(n+1)2]=2C(n+2)3=右边
下面那个同样可以这样解
还有比这简便的方法的吗?

这个就直接用数学归纳法了
要不拆项也可以很容易看出来啊
n(n+1)=n^2+n
n^2的和是(2n+1)(n+1)n/6
n的和是(n+1)*n/2
两式相加就得到上面的结果了
加点分给我得了

因为1+2+…+n=(n+1)*n\2,1*1+2*2+3*3+…+n*n=n*(n+1)*(2*n+1)\6,1^3+2^3+…n^3=(n+1)^2*n^2\4,代入即得,或者直接用数学归纳法。