已知数列{an}满足前N项和sn=n平方+1数列{bn}满足bn=2/an +1且前n项和为Tn 设T 2n+1 -Tn
2个回答
(1)
∵数列{an}满足前N项和sn=n平方+1
∴Sn=n^2+1
S(n-1)=(n-1)^2+1
An=Sn-S(n-1)
=n^2+1-[(n-1)^2+1]
=2n-1
A1=S1=2
Bn=2/An +1=2/(2n-1)+1=(2n+1)/(2n-1)
B1=2/A1+1=2
Bn是一个首项为2,通项为(2n+1)/(2n-1) 的数列
(2)
Cn=T(2n+1)-Tn
要判断Cn的单调性只要判断Cn-C(n-1)是大于0还是小于0即可
Cn-C(n-1)=T(2n+1)-Tn-[T(2n-1)-T(n-1)]
=[T(2n+1)-T(2n-1)]-[Tn-T(n-1)]
=B(2n+1)+B(2n)-Bn
=[2(2n+1)+1]/[2(2n+1)-1]+[2(2n)+1]/[2(2n)-1]-[(2n+1)/(2n-1)]
=1+2[1/(4n+1)+1/(4n-1)-1/(2n-1)]
∵1/(4n+1)+1/(4n-1)-1/(2n-1)
= (1-8n)/[(4n+1)*(4n-1)*(2n-1)]
又∵1-8n0,4n-1>0,2n-1>0
∴(1-8n)/[(4n+1)*(4n-1)*(2n-1)]
相关问题
