Question

是否存在常数a,b,c,使等式1^2+3^2……(2n-1)^2=an(bn^2+c)/3

Answer 1

奇数平方和求和公式为：1^2+3^2……(2n-1)^2=n(4n^2-1)/3

所以,存在a=1,b=4,c=-1

相关证明：

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)