三校生数学,an和sn的关系.数列{an}的首项a1=1,前n项和Sn与通项an满足2S
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三校生数学,an和sn的关系.数列{an}的首项a1=1,前n项和Sn与通项an满足2S
三校生数学,an和sn的关系.
数列{an}的首项a1=1,前n项和Sn与通项an满足2S²n=2anSn-an(n≧2),求数列{an}的通项公式
三校生数学,an和sn的关系.
数列{an}的首项a1=1,前n项和Sn与通项an满足2S²n=2anSn-an(n≧2),求数列{an}的通项公式
2(Sn)^2= 2an.Sn-an
= [Sn- S(n-1)].(2Sn -1 )
= 2(Sn)^2-Sn - 2Sn.S(n-1) + S(n-1)
S(n-1) -Sn = 2Sn.S(n-1)
1/Sn -1/S(n-1) = 2
{1/Sn}是等差数列,d=2
1/Sn -1/S1= 2(n-1)
1/Sn =(4n-3)/2
Sn = 2/(4n-3)
an =Sn -S(n-1)
= 2[ 1/(4n-3) - 1/(4n+1) ]
an =1 ; n=1
=2[ 1/(4n-3) - 1/(4n+1) ] ; n≥2
= [Sn- S(n-1)].(2Sn -1 )
= 2(Sn)^2-Sn - 2Sn.S(n-1) + S(n-1)
S(n-1) -Sn = 2Sn.S(n-1)
1/Sn -1/S(n-1) = 2
{1/Sn}是等差数列,d=2
1/Sn -1/S1= 2(n-1)
1/Sn =(4n-3)/2
Sn = 2/(4n-3)
an =Sn -S(n-1)
= 2[ 1/(4n-3) - 1/(4n+1) ]
an =1 ; n=1
=2[ 1/(4n-3) - 1/(4n+1) ] ; n≥2
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