已知数列{an}满足a1=1,an>0,Sn是数列{an}的前n项和,对任意的n∈N*,有2Sn=2an2+an-1.(1)求数列{an
已知数列{an}满足a1=1,an>0,Sn是数列{an}的前n项和,对任意的n∈N*,有2Sn=2an2+an-1.(1)求数列{an}的通项公式;(2)记bn=an2...
已知数列{an}满足a1=1,an>0,Sn是数列{an}的前n项和,对任意的n∈N*,有2Sn=2an2+an-1.(1)求数列{an}的通项公式;(2)记bn=an2n,求数列{bn}的前n项和Tn.
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(1)由2Sn=2an2+an-1,得2Sn+1=2an+12+an+1?1.
两式相减得:2an+1=2(an+1-an)(an+1+an)+(an+1-an)?
(an+1+an)(2an+1-2an-1)=0
∵an>0,∴2an+1-2an-1=0,∴an+1=an+
.
∴数列{an}是以1为首项,
为公差的等差数列
∴an=
;
( 2 )由bn=
=
则Tn=
+
+
+…+
,①
Tn=
+
+
+…+
.②
①-②得:
Tn=
+
+
+
+…+
?
=
两式相减得:2an+1=2(an+1-an)(an+1+an)+(an+1-an)?
(an+1+an)(2an+1-2an-1)=0
∵an>0,∴2an+1-2an-1=0,∴an+1=an+
1 |
2 |
∴数列{an}是以1为首项,
1 |
2 |
∴an=
n+1 |
2 |
( 2 )由bn=
an |
2n |
n+1 |
2n+1 |
则Tn=
2 |
22 |
3 |
23 |
4 |
24 |
n+1 |
2n+1 |
1 |
2 |
2 |
23 |
3 |
24 |
4 |
25 |
n+1 |
2n+2 |
①-②得:
1 |
2 |
2 |
22 |
1 |
23 |
1 |
24 |
1 |
25 |
1 |
2n+1 |
n+1 |
2n+2 |
=