已知数列{an}中,a1=3,a2=5,其前n项和Sn满足Sn+Sn-2=2Sn-1+2n-1(n≥3),令bn=1anan+1(Ⅰ)求数列{an}的通项公式;(Ⅱ)令Tn=b1+b2•2+b3•22

题目简介

已知数列{an}中,a1=3,a2=5,其前n项和Sn满足Sn+Sn-2=2Sn-1+2n-1(n≥3),令bn=1anan+1(Ⅰ)求数列{an}的通项公式;(Ⅱ)令Tn=b1+b2•2+b3•22

题目详情

已知数列{an}中,a1=3,a2=5,其前n项和Sn满足Sn+Sn-2=2Sn-1+2n-1(n≥3),令bn=
1
anan+1

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)令Tn=b1+b2•2+b3•22+…bn•2n-1
求证:①对于任意正整数n,都有Tn
1
6
.②对于任意的m∈(0,
1
6
)
,均存在n0∈N*,使得n≥n0时,Tn>m.
题型:解答题难度:中档来源:湖北模拟

答案

(Ⅰ)由题意知Sn-Sn-1=Sn-1-Sn-2+2n-1(n≥3),
即an=an-1+2n-1(n≥3)…(1分)
∴an=(an-an-1)+(an-1-an-2)+…+(a3-a2)+a2
=2n-1+2n-2+…+22+5
=2n-1+2n-2+…+22+2+1+2
=2n+1,n≥3.…(3分)
检验知n=1,2时,结论也成立
故an=2n+1.…(4分)
(Ⅱ) ①由于bn2n-1=class="stub"1
(2 n+1)(2n+1 +1)
2n-1

=class="stub"1
2
(2n+1+1)-(2n+1)
(2n+1)(2n+1+1)

=class="stub"1
2
(class="stub"1
2 n+1
-class="stub"1
2n+1+1
)

故Tn=b1+b2•2+b3•22+…+bn•2n-1
=class="stub"1
2
(class="stub"1
1+2
-class="stub"1
1+22
+class="stub"1
1+22
-class="stub"1
1+23
+…+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
1+2

=class="stub"1
6
.…(9分)
②若Tn>m,其中m∈(0,class="stub"1
6
)
,则有class="stub"1
2
(class="stub"1
1+2
-class="stub"1
2n+1+1
)>m

2n+1>class="stub"3
1-6m
-1

n>log2(class="stub"3
1-6m
-1)-1>0

n0=[log2(class="stub"3
1-6m
-1)-1]+1

=[log2(class="stub"3
1-6m
-1)
](其中[x]表示不超过x的最大整数),
则当n>n0时,Tn>m.…(14分)

更多内容推荐