数列an,的前n项和为Sn,2an=Sn+2,n∈Ν*（1）求数列an的通项公式（2）若bn=log₂an+log₂a（n+1）,求数列﹛1/bn(bn+1)﹜的前n项和Tn

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数列an,的前n项和为Sn,2an=Sn+2,n∈Ν*（1）求数列an的通项公式（2）若bn=log₂an+log₂a（n+1）,求数列﹛1/bn(bn+1)﹜的前n项和Tn
数列an,的前n项和为Sn,2an=Sn+2,n∈Ν*
（1）求数列an的通项公式
（2）若bn=log₂an+log₂a（n+1）,求数列﹛1/bn(bn+1)﹜的前n项和Tn

数列an,的前n项和为Sn,2an=Sn+2,n∈Ν*（1）求数列an的通项公式（2）若bn=log₂an+log₂a（n+1）,求数列﹛1/bn(bn+1)﹜的前n项和Tn
(1)
2an =Sn+2
n=1
a1=2
2an =Sn+2
2[Sn-S(n-1)] =Sn+2
Sn +2= 2[S(n-1)+2]
(Sn +2)/[S(n-1)+2] =2
(Sn +2)/(S1 +2)=2^(n-1)
Sn +2 =2^(n+1)
Sn = -2+ 2^(n+1)
an =Sn-S(n-1) = 2^n
(2)
bn = logan + loga(n+1)
= n+(n+1)
= 2n+1
1/[bn.b(n+1)] = 1/[(2n+1)(2n+3)]
=(1/2)(1/(2n+1)-1/(2n+3)]
Tn = (1/2)[ 1/3 -1/(2n+3)]
= n/[3(2n+3)]

　　Sn=2An+2;
n>=2,An=Sn-S(n-1)=2An+2-(2A(n-1)+2);
An/A(n-1)=2;
n=1,A1=2
An=二乘以二的N-1此方

1)令n=n-1，再两式作差，可得2an-2a[n-1]=a[n]，又a1=2所以an=2^n
2)bn=2n加1，然后裂项
Tn=0.5*(1/3-1/5 1/5-1/7..... 1/(2n加1)-1/(2n加3))
整理得：Tn=n/(6n加9)

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