高一数学-关于数列

来源：百度知道 编辑：UC知道 时间：2024/05/15 03:30:41
有一等差数列｛an｝，且有a1+3a2+5a3+…+（2n-1）an=2n+1.求an。

过程详细点！！！

a(n) = a + (n-1)d, n = 1,2,...

(2n-1)a(n) = (2n-1)a + (2n-1)(n-1)d

= 2na - a + 2dn^2 - 3dn + d

= 2dn^2 + n[2a - 3d] + d-a

2n+1 = a(1) + 3a(2) + ... + (2n-1)a(n)

= 2d[1 + 2^2 + ... + n^2] + [2a-3d][1 + 2 + ... + n] + (d-a)n

= 2dn(n+1)(2n+1)/6 + (2a-3d)n(n+1)/2 + (d-a)n

= dn(n+1)(2n+1)/3 + (2a-3d)n(n+1)/2 + (d-a)n

= d[2n^3 + 3n^2 + n]/3 + (2a-3d)[n^2 + n]/2 + (d-a)n

= 2dn^3/3 + n^2[d + (2a-3d)/2] + n[d/3 + (2a-3d)/2 + d-a]

d = 0,
d + (2a-3d)/2 = 0,
a = 0.

a(n) = 0

题目有问题啊。。