Question

是否存在常数a,b,c,使等式3^2+5^2+...+(2n+1)^2=[n(4n^2+an+b)]/3,对于任意正整数

Answer 1

利用1^2+2^2+...+n^2=n(n+1)(2n+1)/6

2^2+4^2+..+(2n)^2

=4(1^2+2^2+...+n^2)

=2n(n+1)(2n+1)/3

3^2+5^2+...+(2n+1)^2

=1^2+2^2+...+(2n+1)^2

-1^2-[2^2+4^2+...+(2n)^2]

=(2n+1)(2n+2)(4n+3)/6-1-2n(n+1)(2n+1)/3

=n(4n^2+12n+11)/3

所以存在且仅有a=12,b=11,使得3^2+5^2+...+(2n+1)^2=[n(4n^2+an+b)]/3成立