题目内容
(2012•宿州三模)已知数列{an}的前n项和Sn满足:Sn=t(Sn-an+1)(t>0),且4a3是a1与2a2的等差中项.
(Ⅰ)求t的值及数列{an}的通项公式;
(Ⅱ)设bn=
,求数列{bn}的前n项和Tn.
(Ⅰ)求t的值及数列{an}的通项公式;
(Ⅱ)设bn=
2n+1 | an |
分析:(Ⅰ)当n≥2时,Sn=t(Sn-an+1),再写一式,两式相减,可得{an}是首项a1=t,公比等于t的等比数列,利用4a3是a1与2a2的等差中项,即可求得结论;
(Ⅱ)由(Ⅰ),得bn=(2n+1)×2n,利用错位相减法,可求数列{bn}的前n项和Tn.
(Ⅱ)由(Ⅰ),得bn=(2n+1)×2n,利用错位相减法,可求数列{bn}的前n项和Tn.
解答:解:(Ⅰ)当n=1时,S1=t(S1-a1+1),所以a1=t,
当n≥2时,Sn=t(Sn-an+1)①
Sn-1=t(Sn-1-an-1+1),②
①-②,得an=t•an-1,即
=t.
故{an}是首项a1=t,公比等于t的等比数列,所以an=tn,…(4分)
故a2=t2,a3=t3
由4a3是a1与2a2的等差中项,可得8a3=a1+2a2,即8t3=t+2t2,
因t>0,整理得8t2-2t-1=0,解得t=
或t=-
(舍去),
所以t=
,故an=
.…(6分)
(Ⅱ)由(Ⅰ),得bn=
=(2n+1)×2n,
所以Tn=3×2+5×22+7×23+…+(2n-1)×2n-1+(2n+1)×2n,③
2Tn=3×22+5×23+7×24+…+(2n-1)×2n+(2n+1)×2n+1,④
③-④,得-Tn=3×2+2(22+23+…+2n)-(2n+1)×2n+1 …(8分)
=-2+2n+2-(2n+1)×2n+1=-2-(2n-1)×2n+1…(11分)
所以Tn=2+(2n-1)×2n+1.…(12分)
当n≥2时,Sn=t(Sn-an+1)①
Sn-1=t(Sn-1-an-1+1),②
①-②,得an=t•an-1,即
an |
an-1 |
故{an}是首项a1=t,公比等于t的等比数列,所以an=tn,…(4分)
故a2=t2,a3=t3
由4a3是a1与2a2的等差中项,可得8a3=a1+2a2,即8t3=t+2t2,
因t>0,整理得8t2-2t-1=0,解得t=
1 |
2 |
1 |
4 |
所以t=
1 |
2 |
1 |
2n |
(Ⅱ)由(Ⅰ),得bn=
2n+1 |
an |
所以Tn=3×2+5×22+7×23+…+(2n-1)×2n-1+(2n+1)×2n,③
2Tn=3×22+5×23+7×24+…+(2n-1)×2n+(2n+1)×2n+1,④
③-④,得-Tn=3×2+2(22+23+…+2n)-(2n+1)×2n+1 …(8分)
=-2+2n+2-(2n+1)×2n+1=-2-(2n-1)×2n+1…(11分)
所以Tn=2+(2n-1)×2n+1.…(12分)
点评:本题考查数列递推式,考查数列的通项,考查错位相减法求数列的和,确定数列为等比数列是关键.
练习册系列答案
相关题目