题目

已知an=n^2,抽去数列的第一项第三项...第（3n-2）项,设此时的数列为dn,求dn的前n项和sn.

提问时间：2020-11-03

答案

a(n) = n^2,
a(3n-2) = (3n-2)^2

d(2n-1) = a(3n-1)= (3n-1)^2 = 9n^2 - 6n + 1
d(2n) = a(3n) = (3n)^2 = 9n^2,
d(2n-1) + d(2n) = 18n^2 - 6n + 1,
s(2n) = [d(1)+d(2)] + [d(3)+d(4)] + ... + [d(2n-1)+d(2n)]
= 18n(n+1)(2n+1)/6 - 6n(n+1)/2 + n
= 3n(n+1)(2n+1) - 3n(n+1) + n
= 3n(n+1)*2n + n
= 6(n+1)n^2 + n
= n(6n^2+6n+1)
s(2n-1) = s(2n) - d(2n) = 6(n+1)n^2 + n - 9n^2 = 6n^2 + 6n^3 + n - 9n^2
= 6n^3 - 3n^2 + n
=n(6n^2 - 3n+1)