1x2+2x3+3x4+...+100x101=? 1x2+2x3+3x4+...+n(n+1)=

Question

如题所述

Answer 1

1x2+2x3+3x4+...+100x101
=[（1x2+2x3+3x4+...+100x101）x3]/3
=[1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+....+100x101x(102-109)]/3
=(-0x1x2+1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-....-109x100x101+100x101x102)/3
=100x101x102/3
=343400
1x2+2x3+3x4+...+n(n+1)
={[1x2+2x3+3x4+...+nx(n+1）]x3}/3
={1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+....+nx(n+1)x[(n+2)-(n-1)]/3
=[-0x1x2+1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-....-(n-1)xnx(n+1)+nx(n+1)x(n+2)]/3
=nx(n+1)x(n+2)/3

Answer 2

1*2 = (（1+2）^2 - 1^2-2^2)/2
2*3 = (（2+3）^2 - 2^2-3^2)/2
3*4 = (（3+4）^2 - 3^2-4^2)/2

前n个奇数平方和的公式：1^2+..+(2n-1)^2=(1/3)n(4n^2-1)
过程 ：
1^2+2^2+...+n^2=n(n+1)(2n+1)/6
1^2+2^2+...+(2n)^2=2n(2n+1)(4n+1)/6=n(2n+1)(4n+1)/3
2^2+4^2+...+(2n)^2=4(1^2+2^2+...+n^2)=4n(n+1)(2n+1)/6=2n(n+1)(2n+1)/3
1^2+3^2+...(2n-1)^2=[1^2+2^2+...+(2n)^2]-[2^2+4^2+...+(2n)^2]
=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3=n(2n+1)(2n-1)/3=(1/3)n(4n^2-1)

知道这些就可以算出来了。

参考资料：http://zhidao.baidu.com/question/297571261.html