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

n为正整数的情况下,1^2+3^2+5^2+……+（2n-1）^2=sigma（4n^2-4n+1)=4(1^2+2^2+3^3+…+n^2)-4(1+2+3+…+n)+n注意,重点在于：首先,1^2+2^2+3^2+…+n^2=[n*(n+1)*(2n+1)]/6=1/3n^3+1/2n^2+1/6n；其次,1+2+3+…+n＝1/2n^2+1/2n故1^2+3^2+5^2+……+（2n-1）^2＝4（1/3n^3+1/2n^2+1/6n）-4(1/2n^2+1/2n)+n=……展开,合并同类项……=4/3n^3-1/3n=1/3n(4n^2-1)=1/3an(bn^2+c)然后就abc就“一一对应”吧。会了吗?