数列an,的前n项和为Sn,2an=Sn+2,n∈Ν*
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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
(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*](/uploads/image/z/6819608-56-8.jpg?t=%E6%95%B0%E5%88%97an%2C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C2an%3DSn%2B2%2Cn%E2%88%88%CE%9D%2A)
(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)]
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)]
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