已知数列{an}的前n项和Sn=n2.(I)求数列{an}的通项公式;(II)设an=2nbn,求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}的前n项和Sn=n2.(I)求数列{an}的通项公式;(II)设an=2nbn,求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}的前n项和Sn=n2
(I)求数列{an}的通项公式;
(II)设an=2nbn,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:唐山二模

答案

(I)当n=1时,a1=S1=1,
当n≥2时,an=Sn-Sn-1=h2-(n-1)2=2n-1,
且对n=1成成立.
∴an=2n-1.
(II)由an=2nbn=2n-1,得bn=class="stub"2n-1
2n
Tn=class="stub"1
2
+class="stub"3
22
+class="stub"5
23
+…+class="stub"2n-1
2n
,①2Tn=1+class="stub"3
2
+class="stub"5
22
+…+class="stub"2n-1
2n-1
,②
②-①,得Tn=1+1+class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n-2
-class="stub"2n-1
2n
=3-class="stub"2n+3
2n

更多内容推荐