已知函数f(x)=(x-1)2,数列{an}是各项均不为0的等差数列,点(an+1,S2n-1)在函数f(x)的图象上;数列{bn}满足bn=(34)n-1.(I)求an;(II)若数列{cn}满足c

题目简介

已知函数f(x)=(x-1)2,数列{an}是各项均不为0的等差数列,点(an+1,S2n-1)在函数f(x)的图象上;数列{bn}满足bn=(34)n-1.(I)求an;(II)若数列{cn}满足c

题目详情

已知函数f(x)=(x-1)2,数列{an}是各项均不为0的等差数列,点(an+1,S2n-1)在函数f(x)的图象上;数列{bn}满足bn=(
3
4
)n-1

(I)求an
(II)若数列{cn}满足cn=
an
4n-1bn
,证明:c1+c2+c3+…+cn<3.
题型:解答题难度:中档来源:潍坊二模

答案

(I)因为点(an+1,S2n-1)在函数f(x)的图象上,所以an2=S2n-1,
令n=1,2得
a21
=S1
a22
=S3
,即
a21
=a1,…①
a22
=3a1+3d,…②
解得
a1=1
d=2

∴an=2n-1;
(II)由(I)得cn=
an
4n-1bn
=class="stub"2n-1
4n-1•(class="stub"3
4
)n-1
=class="stub"2n-1
3n-1

令Tn=c1+c2+c3+…+cn,
则Tn=class="stub"1
30
+class="stub"3
31
+class="stub"5
32
+…+class="stub"2n-3
3n-2
+class="stub"2n-1
3n-1
,①
class="stub"1
3
Tn=class="stub"1
31
+class="stub"3
32
+class="stub"5
33
+…+class="stub"2n-3
3n-1
+class="stub"2n-1
3n
,②
①-②得class="stub"2
3
Tn=class="stub"1
3 
+class="stub"2
32
+class="stub"2
33
+…+class="stub"2
3n-1
-class="stub"2n-1
3n
=1+class="stub"2
3
×
1-class="stub"1
3n-1
1-class="stub"1
3
-class="stub"2n-1
3n
=2-
2(n+1)
3n

∴Tn=3-class="stub"n+1
3n-1
<3.

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