数列{an}的前n项的和为Sn,对于任意的自然数an>0,4Sn=(an+1)2(Ⅰ)求证:数列{an}是等差数列,并求通项公式(Ⅱ)设bn=an3n,求和Tn=b1+b2+…+bn.-数学

题目简介

数列{an}的前n项的和为Sn,对于任意的自然数an>0,4Sn=(an+1)2(Ⅰ)求证:数列{an}是等差数列,并求通项公式(Ⅱ)设bn=an3n,求和Tn=b1+b2+…+bn.-数学

题目详情

数列{an}的前n项的和为Sn,对于任意的自然数an>0,4Sn=(an+1)2
(Ⅰ)求证:数列{an}是等差数列,并求通项公式
(Ⅱ)设bn=
an
3n
,求和Tn=b1+b2+…+bn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)证明:∵4S1=4a1=(a1+1)2,∴a1=1.
当n≥2时,4an=4Sn-4Sn-1=(an+1)2-(an-1+1)2,
∴2(an+an-1)=an2-an-12,
又{an}各项均为正数,∴an-an-1=2,
∴数列{an}是等差数列,
∴an=2n-1;
(Ⅱ)bn=
an
3n
=class="stub"2n-1
3n

∴Tn=b1+b2+…+bn=class="stub"1
31
+class="stub"3
32
+…+class="stub"2n-1
3n
---①
class="stub"1
3
Tn=class="stub"1
32
+class="stub"3
33
+…+class="stub"2n-3
3n
+class="stub"2n-1
3n+1
---②
①-②class="stub"2
3
Tn=class="stub"1
31
+2(class="stub"1
32
+class="stub"1
33
+…+class="stub"1
3n
)-class="stub"2n-1
3n+1
=class="stub"2
3
-class="stub"2n+2
3n+1

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

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