数列{an}中,a1=3,Sn为其前n项的和,满足Sn=Sn-1+an-1+2n-1(n≥2),令bn=1anan+1(1)写出数列{an}的前四项,并求数列{an}的通项公式(2)若f(x)=2x-

题目简介

数列{an}中,a1=3,Sn为其前n项的和,满足Sn=Sn-1+an-1+2n-1(n≥2),令bn=1anan+1(1)写出数列{an}的前四项,并求数列{an}的通项公式(2)若f(x)=2x-

题目详情

数列{an}中,a1=3,Sn为其前n项的和,满足Sn=Sn-1+an-1+2n-1(n≥2),令bn=
1
anan+1

(1)写出数列{an}的前四项,并求数列{an}的通项公式
(2)若f(x)=2x-1,求和:b1f(1)+b2f•(2)+…+bnf(n)
(3)设cn=
n
an
,求证:数列{cn}的前n项和Qn<2.
题型:解答题难度:中档来源:汕头模拟

答案

(1)数列的前四项:a1=3,a2=5,a3=9,a4=17(2分)
Sn=Sn-1+an-1+2n-1(n≥2)⇒an=an-1+2n-1(n≥2)(3分)
当n≥2时,an=(an-an-1)+•+(a2-a1)+a1=2n-1••+2n-2++22+2•+3=2n+1
经验证a1也符合,所以an=2n.+1(5分)
(2)bnf(n)=
2n-1
(2n+1)(2n+1+1)
=class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+1+1
)
,(7分)
∴b1f(1)+b2f(•2)+…+bnf(n)=class="stub"1
2
(class="stub"1
2+1
-class="stub"1
22+1
)+class="stub"1
2
(class="stub"1
22+1
-class="stub"1
23+1
)+class="stub"1
2
(class="stub"1
23+1
-class="stub"1
24+1
)
+…+class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+1+1
)
=class="stub"1
2
(class="stub"1
2+1
-class="stub"1
2n+1+1
)=class="stub"1
6
-class="stub"1
2n+2+2
(9分)
(3)由cn=class="stub"n
2n-1
<class="stub"n
2n

Qn=class="stub"1
2+1
+class="stub"1
22+1
+…+class="stub"1
2n+1
<class="stub"1
2
+class="stub"2
22
+…+class="stub"n
2n
(11分)
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"3
24
+…+class="stub"n
2n+1

相减,得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
=
class="stub"1
2
×(1-class="stub"1
2n
)
1-class="stub"1
2
-class="stub"n
2n+1
=1-class="stub"1
2n
-class="stub"n
2n+1

所以Tn=2-class="stub"n+2
2n

所以Qn<class="stub"1
2
+class="stub"2
23
+…+class="stub"n
2n
=2-class="stub"n+2
2n
<2
(14分)

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