题目

已知 1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6,则1^2+3^2+...+（2n+1）^2=?

提问时间：2020-11-06

答案

2^2+4^2+...+(2n)^2
=2^2(1^2+2^2+..+n^2)
= 4n(n+1)(2n+1)/6
1^2+2^2+...+(2n+1)^2 = (2n+1)(2n+2)(2n+3)/6
1^2+3^2+...+（2n+1）^2
= 1^2+2^2+...+(2n+1)^2 - ( 2^2+4^2+...+(2n)^2 )
=(2n+1)(2n+2)(2n+3)/6 - 4n(n+1)(2n+1)/6
=[(n+1)(2n+1)/3 ] ( (2n+3) - 2n )
= (n+1)(2n+1)