问题
解答题
已知等差数列{an}的前n项和为Sn,且a2=1,S11=33. (1)求{an}的通项公式; (2)设bn=(
|
答案
(1)由已知,且a2=a1+d=1,S11=11a1+55d=33,解得a1=
,d=1 2
,an=1 2
.n 2
(2)bn=(
)an=(1 4
) n,1 2
=bn+1 bn
,数列bn}是以1 2
为公比的等比数列.1 2
an•bn=
.(n 2
)n=n•(1 2
)n+1 1 2
Tn=1×(
)2+2×(1 2
)3+…+n•(1 2
)n+1 ①1 2
Tn=+1×(1 2
)3+2(1 2
)4+…+(n-1)•(1 2
)n+1+…+n•(1 2
)n+2 ②1 2
②-①得
Tn=(1 2
)2+(1 2
)3+(1 2
)4…+(1 2
)n+1-n•(1 2
)n+2 1 2
=
-n•(
[1-(1 4
)n]1 2 1- 1 2
)n+21 2
∴Tn=1-(
)n-n•(1 2
)n+1=1-1 2
.2-n 2n+1