设{an}是正数组成的数列,其前n项和为Sn,并且 - 知识问答

设{an}是正数组成的数列,其前n项和为Sn,并且对于所有的n属于N*,都有8Sn=(an+2)^2

1个回答

8Sn=(an+2)^2 (1)
n=1
8a1=(a1+2)^2
(a1)^2-4a1+4=0
a1=2
8S(n-1)=(a(n-1)+2)^2 (2)
(1)-(2)
8an=(an+2)^2 -(a(n-1)+2)^2
(an)^2- [a(n-1)]^2 - 4[an+a(n-1)] =0
[an+a(n-1)] .[an-a(n-1)-4]=0
an-a(n-1)-4=0
an-a1=4(n-1)
an =4n-2= 2(2n-1)
bn=4/[an.a(n+1)]
=1/[(2n-1)(2n+1)]
= (1/2)[ 1/(2n-1)- 1/(2n+1) ]
Tn=b1+b2+...+bn
=(1/2)[ 1- 1/(2n+1)]
Tn

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