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.
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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.
![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.](/uploads/image/z/4462430-14-0.jpg?t=%E5%B7%B2%E7%9F%A5an%3Dn%5E2%2C%E6%8A%BD%E5%8E%BB%E6%95%B0%E5%88%97%E7%9A%84%E7%AC%AC%E4%B8%80%E9%A1%B9%E7%AC%AC%E4%B8%89%E9%A1%B9...%E7%AC%AC%EF%BC%883n-2%EF%BC%89%E9%A1%B9%2C%E8%AE%BE%E6%AD%A4%E6%97%B6%E7%9A%84%E6%95%B0%E5%88%97%E4%B8%BAdn%2C%E6%B1%82dn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn.)
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)
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)
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