已知函数f(x)=3-x,等比数列an的前n项和为f(n)-c,正项数列bn的首项为c,且前n项和Sn满足Sn-Sn=Sn-1+Sn-1,(n≥2)(1)求c,并求数列{an}和{bn}的通项公式;(

题目简介

已知函数f(x)=3-x,等比数列an的前n项和为f(n)-c,正项数列bn的首项为c,且前n项和Sn满足Sn-Sn=Sn-1+Sn-1,(n≥2)(1)求c,并求数列{an}和{bn}的通项公式;(

题目详情

已知函数f(x)=3-x,等比数列an的前n项和为f(n)-c,正项数列bn的首项为c,且前n项和Sn满足Sn-
Sn
=Sn-1+
Sn-1
,(n≥2)

(1)求c,并求数列{an}和{bn}的通项公式;
(2)求数列{bn(1-
1
2
an)}
的前n项和为Tn
题型:解答题难度:中档来源:不详

答案

(1)∵等比数列an的前n项和为f(n)-c,
∴a1=f(1)-c=class="stub"1
3
-c,
∴a2=[f(2)-c]-[f(1)-c]=-class="stub"2
9
,a3=[f(3)-c]-[f(2)-c]=-class="stub"2
27

又数列{an}成等比数列,
a1=
a22
a3
=-class="stub"2
3

∵a1=class="stub"1
3
-c
∴-class="stub"2
3
=class="stub"1
3
-c,∴c=1
又公比q=
a2
a1
=class="stub"1
3

所以an=-class="stub"2
3
(class="stub"1
3
)
n-1
,n∈N;
∵Sn-Sn-1=(
Sn-Sn-1
)(
Sn
+
Sn-1
)
=
Sn
+
Sn-1
(n≥2)
又bn>0,
Sn
>0,∴
Sn
-
Sn-1
=1;
∴数列{
Sn
}构成一个首项为1公差为1的等差数列,
Sn
=1+(n-1)×1=n,Sn=n2
当n≥2,bn=Sn-Sn-1=n2-(n-1)2=2n-1;
又b1=c=1适合上式,∴bn=2n-1(n∈N);
(2)由(1)知bn(1-class="stub"1
2
an)
=(2n-1)+(2n-1)•(class="stub"1
3
)n
设(2n-1)•(class="stub"1
3
)n前n项和为Qn   设数列2n-1的前n项和为Sn
Qn=class="stub"1
3
+3×(class="stub"1
3
)2+5×(class="stub"1
3
)3+…+(2n-3)•(class="stub"1
3
)n-1+(2n-1)•(class="stub"1
3
)n     ①
class="stub"1
3
Qn=(class="stub"1
3
)2+3×(class="stub"1
3
)3+5×(class="stub"1
3
)4+…+(2n-3)•(class="stub"1
3
)n+(2n-1)•(class="stub"1
3
)n+1  ②
①-②得:class="stub"2
3
QN=class="stub"1
3
+2[(class="stub"1
3
)2+(class="stub"1
3
)3+(class="stub"1
3
)4++(class="stub"1
3
)n]-(2n-1)(class="stub"1
3
)n+1
=class="stub"2
3
-(2n+2)(class="stub"1
3
)n+1

∴Qn=1-(n+1)(class="stub"1
3
)n
∴Sn=n2
∴Tn=Sn+Qn=n2+1-(n+1)(class="stub"1
3
)n

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