设数列{an}的前n项和为Sn=2an-2n(Ⅰ)求a1,a2(Ⅱ)设cn=an+1-2an,证明:数列{cn}是等比数列(Ⅲ)求数列{n+12cn}的前n项和为Tn.-高二数学

题目简介

设数列{an}的前n项和为Sn=2an-2n(Ⅰ)求a1,a2(Ⅱ)设cn=an+1-2an,证明:数列{cn}是等比数列(Ⅲ)求数列{n+12cn}的前n项和为Tn.-高二数学

题目详情

设数列{an}的前n项和为Sn=2an-2n
(Ⅰ)求a1,a2
(Ⅱ)设cn=an+1-2an,证明:数列{cn}是等比数列
(Ⅲ)求数列{
n+1
2cn
}
的前n项和为Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵a1=S1,2a1=S1+2,
∴a1=2,S1=2,
由2an=Sn+2n知,2an+1=Sn+1+2n+1=an+1+Sn+2n+1
得an+1=sn+2n+1①,
∴a2=S1+22=2+22=6;
(Ⅱ)由题设和①式知an+1-2an=(Sn+2n+1)-(Sn+2n)=2n+1-2n=2n,
即cn=2n,
cn+1
cn
=2(常数),
∴{cn}是首项为2,公比为2的等比数列.
(Ⅲ)∵cn=an+1-2an=2n,
class="stub"n+1
2cn
=class="stub"n+1
2n+1

∴数列{class="stub"n+1
2cn
}
的前n项和Tn=class="stub"2
22
+class="stub"3
23
+class="stub"4
24
+…+class="stub"n+1
2n+1

class="stub"1
2
Tn=class="stub"2
23
++class="stub"4
24
+…+class="stub"n
2n+1
+class="stub"n+1
2n-2

相减得class="stub"1
2
Tn=class="stub"2
22
+class="stub"1
23
+class="stub"1
24
+class="stub"1
25
…+class="stub"n
2n+1
-class="stub"n+1
2n+2
=class="stub"1
2
+
class="stub"1
23
×(1-class="stub"1
2n-1
)
1-class="stub"1
2
-class="stub"n+1
2n+2
=class="stub"3
4
-class="stub"1
2n+1
-class="stub"n+1
2n+2

∴Tn=class="stub"3
2
-class="stub"1
2n
-class="stub"n+1
2n+1

更多内容推荐