数列a1=3,数列满足2an=Sn乘以S(n-1),求通项公式.
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数列a1=3,数列满足2an=Sn乘以S(n-1),求通项公式.
数列a1=3,数列满足2an=Sn乘以S(n-1),求通项公式.
数列a1=3,数列满足2an=Sn乘以S(n-1),求通项公式.
因为an=Sn-S(n-1) n>=2
2an=Sn*S( n-1)
2(Sn-S(n-1))=Sn*S(n-1) 两边除以Sn*S(n-1)
1/Sn-1/S(n-1)= -1/2
所以数列{1/Sn}是以-1/2为公差 1/S1=1/3为首项的等差数列
1/Sn=-n/2 + 5/6
Sn=1/(-n/2 +5/6)
an=Sn-S(n-1)=1/(-n/2 +5/6)- 1/(-n/2+4/3) =6/(5-3n)-6/(8-3n) n>=2
a1=3 不满足an
所以 an =6/(5-3n)-6/(8-3n) n>=2
=3 n=1
2[Sn-S(n-1)]=Sn*S(n-1),所以 1/S(n-1)-1/Sn=1/2,
1/S1=1/a1=1/3,1/Sn=(n-1)/2+1/3=(3n-1)/6,
所以Sn=6/(3n-1),S(n-1)=6/(3n-4) (n>=2),
所以an=1/2*Sn*S(n-1)=18/[(3n-1)(3n-4)] (n>=2),
a1=3不符...
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2[Sn-S(n-1)]=Sn*S(n-1),所以 1/S(n-1)-1/Sn=1/2,
1/S1=1/a1=1/3,1/Sn=(n-1)/2+1/3=(3n-1)/6,
所以Sn=6/(3n-1),S(n-1)=6/(3n-4) (n>=2),
所以an=1/2*Sn*S(n-1)=18/[(3n-1)(3n-4)] (n>=2),
a1=3不符合上式;
所以an=1/2*Sn*S(n-1)=18/[(3n-1)(3n-4)] (n>=2),
an=3 (n=1),
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