已知数列{an}中,a1=3,a2=5,Sn为其前n项和,且满足Sn+Sn-2=2Sn-1+2n-1(n≥3,n∈N*).(1)求数列{an}的通项公式;(2)令bn=nan-1,求数列{bn}的前n

题目简介

已知数列{an}中,a1=3,a2=5,Sn为其前n项和,且满足Sn+Sn-2=2Sn-1+2n-1(n≥3,n∈N*).(1)求数列{an}的通项公式;(2)令bn=nan-1,求数列{bn}的前n

题目详情

已知数列{an}中,a1=3,a2=5,Sn为其前n项和,且满足Sn+Sn-2=2Sn-1+2n-1(n≥3,n∈N*).
(1)求数列{an}的通项公式;
(2)令bn=
n
an-1
,求数列{bn}的前n项和Tn
(3)若f(x)=2x-1,cn=
1
anan+1
,Qn=c1f(1)+c2f(2)+…+cnf(n),求证Qn
1
6
(n∈N*).
题型:解答题难度:中档来源:不详

答案

(1)由Sn+Sn-2=2Sn-1+2n-1an=an-1+2n-1(n≥3,n∈N*)
∵a2=5,∴当n≥3时,an=a2+(a3-a2)+(a4-a3)+…+(an-an-1)=5+22+23+…+2n-1=2n+1,
经验证a1=3,a2=5也符合上式,
an=2n+1(n∈N*)
(2)由(1)可得bn=class="stub"n
an-1
=class="stub"n
2n

Tn=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n
2n
⇒class="stub"1
2
Tn=class="stub"1
22
+class="stub"2
23
+…+class="stub"n-1
2n
+class="stub"n
2n+1
②,
①-②有:class="stub"1
2
Tn=class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
-class="stub"n
2n+1
=1-class="stub"1
2n
-class="stub"n
2n+1

Tn=2-class="stub"n+2
2n

(3)∵f(x)=2x-1cn=class="stub"1
anan+1

cnf(n)=
2n-1
(2n+1)(2n+1+1)
=class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+1+1
)(n∈N*)

∴Qn=c1f(1)+c2f(2)+…+cnf(n)
=class="stub"1
2
[(class="stub"1
21+1
-class="stub"1
22+1
)+(class="stub"1
22+1
-class="stub"1
23+1
)+…+(class="stub"1
2n+1
-class="stub"1
2n+1+1
)]

=class="stub"1
2
(class="stub"1
1+2
-class="stub"1
2n+1+1
)<class="stub"1
2
×class="stub"1
3
=class="stub"1
6

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