是否存在常数a,b,c,是等式1^2+3^2+5^2+...+(2n-1)^2=an/3(bn^2+c)对任意正整数n都

因为1^2+2^2+……+n^2=n(n+1)(2n+1)/6
所以1^2+2^2+……+(2n)^2=(2n)(2n+1)(2*2n+1)/6=n(2n+1)(4n+1)/3
(2^2-1^2)+(4^2-3^2)+……+[(2n)^2-(2n-1)^2]
=(2+1)(2-1)+(4+3)(4-3)+……+(2n+2n-1)(2n-2n+1)
=1+2+……+2n
=2n(2n+1)/2
=n(2n+1)
所以[2^2+4^2+……+(2n)^2]-[1^2+3^3+……+(2n-1)^2]=n(2n+1)
[2^2+4^2+……+(2n)^2]+[1^2+3^3+……+(2n-1)^2]=1^2+2^2+……+(2n)^2=n(2n+1)(4n+1)
两式相减除2
所以
1^2+3^3+……+(2n-1)^2=[n(2n+1)(4n+1)/3-n(2n+1)]/2=n(2n+1)(2n-1)/3
=(n/3)*(4n^2-1)
所以a=1,b=4,c=-1

1^2+3^2+5^2+...+(2n-1)^2=n^2(3n^2-1)为n的四次项
an/3(bn^2+c)不为n的四次项
所以不存在。