巧算 1x2+2x3+3x4+4x5……+19x20 求过程+答案

Question

如题所述

Answer 1

考察一般项：n(n+1)=n^2+n
1×2+2×3+...+19×20
=(1^2+2^2+...+19^2)+(1+2+...+19)
=19×20×39/6 +19×20/2
=2470+190
=2660

一般的：
1×2+2×3+...+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)/6](2n+1+3)
=n(n+1)(n+2)/3本回答被网友采纳

Answer 2

=2x(2-1)+3x(3-1)+4x(4-1)+.........+20x(20-1)
=2^2+3^2+4^2+.........+20^2-(2+3+4+.........+20)
因为1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
所以=2869-209
=2660
谢谢

Answer 3

(1^2+2^2+3^2+......n^2)=1/6*n*(n+1)*(2n+1)
1+2+3+……+n=n*(n+1)/2
原式=(1^2+2^2+3^2+……+19^2)+(1+2+……+19)
=19*20*39/6+19*20/2
=2660