已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=(14)an,求数列{bn}的前n项和Tn.-数学

题目简介

已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.(Ⅰ)求数列{an}的通项公式;(Ⅱ)设bn=(14)an,求数列{bn}的前n项和Tn.-数学

题目详情

已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)设bn=(
1
4
)an
,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)设数列{bn}的公差为d,依题意,得
a1+d=1
11a1+11×5d=33
解得a1=class="stub"1
2
,d=class="stub"1
2

an=a1+(n-1)d=class="stub"n
2

(Ⅱ)bn=(class="stub"1
4
)an=(class="stub"1
2
)n

bn+1÷bn=class="stub"1
2
,故此数列为以class="stub"1
2
为首项和公比的等比数列
Tn=
class="stub"1
2
[1-(class="stub"1
2
)
n
]
1-class="stub"1
2
=1-(class="stub"1
2
)n,n∈N+

更多内容推荐