设各项为正的数列{an}的前几项和Sn满足Sn=0.25(an+1)的平方

来源：百度知道 编辑：UC知道 时间：2024/06/29 02:09:56
设各项为正的数列{an}的前几项和Sn满足Sn=0.25(an+1)的平方，
求证an=2n-1;
设bn=1/an*(an+1),记数列{bn}的前几项和为Ta,求Ta

a(1) = S(1) = 0.25[a(1) + 1]^2,

0 = [a(1) + 1]^2 - 4a(1) = [a(1) - 1]^2,

a(1) = 1.

n >= 2时，
a(n) = S(n) - S(n-1) = 0.25[a(n) + 1]^2 - 0.25[a(n-1) + 1]^2,

[a(n-1) + 1]^2 = [a(n) + 1]^2 - 4a(n) = [a(n) - 1]^2,

a(n-1) + 1 = |a(n) - 1|

若0 < a(n) <= 1，则
a(n-1) + 1 = 1 - a(n),
a(n-1) = -a(n) < 0与{a(n)}为各项为正的数列矛盾。

所以，n >= 2时，
a(n) > 1，
a(n-1) + 1 = a(n) - 1,

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

{a(n)}是首项为a(1) = 1,公差为2的等差数列。
a(n) = 1 + 2(n-1) = 2n - 1, n = 1,2,...

b(n) = 1/[a(n+1)a(n)] = 1/[(2n-1)(2n+1)] = [1/(2n-1) - 1/(2n+1)]/2, n = 1,2,...

T(n) = b(1) + b(2) + ...+ b(n-1) + b(n)

= [1/1 - 1/3]/2 + [1/3 - 1/5]/2 + ... + [1/(2n-3) - 1/(2n-1)]/2 + [1/(2n-1) - 1/(2n+1)]/2

= [1/1 - 1/(2n+1)]/2

= n/(2n+1)