数列题,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属于N,有2Sn=p（2an²+an-1）,p为常数1,求数列｛an｝的通项公式2,记bn=an/2^n,求数列｛bn｝的前n项和Tn

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数列题,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属于N,有2Sn=p（2an²+an-1）,p为常数1,求数列｛an｝的通项公式2,记bn=an/2^

数列题,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属于N,有2Sn=p（2an²+an-1）,p为常数1,求数列｛an｝的通项公式2,记bn=an/2^n,求数列｛bn｝的前n项和Tn
数列题,
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属于N,有2Sn=p（2an²+an-1）,p为常数
1,求数列｛an｝的通项公式
2,记bn=an/2^n,求数列｛bn｝的前n项和Tn

数列题,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属于N,有2Sn=p（2an²+an-1）,p为常数1,求数列｛an｝的通项公式2,记bn=an/2^n,求数列｛bn｝的前n项和Tn
（1）因p为常数,a1=1,故当n=1时,2Sn=2a1=2=p(2*1+1-1)=2p,所以,p=1.
Sn=n(an+a1)/2=n(an+1)/2
2Sn=n(an+1)=2an²+an-1,且,an>0,所以,
an=（n+1）/2
(2).bn=an/2^n=(n+1)/2^(n+1)
Tn=b1+b2+……+bn
=2/2^2+3/2^3+……+n/2^n+(n+1)/2^(n+1)
2Tn=2/2+3/2^2+4/2^3+……+(n+1)/2^n
Tn=2Tn-Tn
= 2/2+1/2^2+1/2^3+……+1/2^n-(n+1)/2^(n+1)
=1/2+(1/2+1/2^2+……+1/2^n)-(n+1)/2^(n+1)
=5/2-(n+5)/2^(n+1)

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