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](/uploads/image/z/4050556-52-6.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%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%8B%A5bn%3Dlog%26%238322%3Ban%2Blog%26%238322%3Ba%EF%BC%88n%2B1%EF%BC%89%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%B9%9B1%2Fbn%28bn%2B1%29%EF%B9%9C%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
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)