连续奇数平方和怎么算?
问题描述：
连续奇数平方和怎么算?
答
1^2+..+(2n-1)^2=(1/3)n(4n^2-1) 证明过程如下: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)1/6 n (1 + n) (2 + n)