正项数列an的前n项和Sn满足Sn^2-（n^2+n-1）Sn-(n^2+n)=0令bn=(n+1)/(n+2)^2an^2其前n_数学解答

正项数列an的前n项和Sn满足Sn^2-（n^2+n-1）Sn-(n^2+n)=0令bn=(n+1)/(n+2)^2an^2其前n项和为Tn
试证明：对于任意的x∈N+都有Tn<5/64

人气：363 ℃　时间：2019-12-03 05:26:14

解答

[Sn - (n^2 + n)](Sn + 1) = 0
因为an 是正项数列 Sn = n^2 + n
an = Sn - Sn-1 = 2n
bn = (n + 1)/4n^2(n+2)^2 = 1/16 * [ 1/n^2 - 1/(n + 2)^2 ]
Tn = 1/16 *
( 1 - 1/9
+ 1/4 - 1/16
+ 1/9 - 1/25
.
+ 1/(n-1)^2 - 1/(n + 1)^2
+ 1/n^2 - 1/(n+2)^2 )
=1/16 * [ 1 + 1/4 -1/(n + 1)^2 - 1/(n+2)^2 ]