首先要利用连续平方和公式:
1^2+2^2+…+n^2
=n(n+1)(2n+1)/6
则连续偶数平方和公式:
2^2+4^2+…+(2n)^2
=2^2*n(n+1)(2n+1)/6
=2n(n+1)(2n+1)/3
则联系奇数平方和公式:
1^2+3^2+…+(2n+1)^2
=(2n+1)(2n+2)(4n+3)/6-
2n(n+1)(2n+1)/3
=(n+1)(2n+1)(2n+3)/3
19=2*9+1
取n=9,则所求和为:
(9+1)(2*9+1)(2*9+3)/3
=10*19*21/3
=1330
附Excel图片验证: