数列{An}中,A1=1,n>1,2Sn=2AnSn-An,求An
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数列{An}中,A1=1,n>1,2Sn=2AnSn-An,求An
数列{An}中,A1=1,n>1,2SnSn=2AnSn-An,求An
数列{An}中,A1=1,n>1,2SnSn=2AnSn-An,求An
2(Sn)^2=2anSn-an
2(Sn)^2-2anSn=-an
2Sn(Sn-an)=-an
2Sn*S(n-1)=-an
2Sn*S(n-1)=-[(Sn-S(n-1)]
2=-1/S(n-1)+1/Sn
1/Sn -1/S(n-1)=2
所以{1/Sn}是等差数列.
S1=a1=1
1/Sn=1/a1+(n-1)d
1/Sn=1/1+(n-1)*2
1/Sn=2n-1
Sn=1/(2n-1)
S(n-1)=1/[2(n-1)-1]
S(n-1)=1/(2n-3)
an=Sn-S(n-1)
=1/(2n-1)-1/(2n-3)
2(Sn)^2-2anSn=-an
2Sn(Sn-an)=-an
2Sn*S(n-1)=-an
2Sn*S(n-1)=-[(Sn-S(n-1)]
2=-1/S(n-1)+1/Sn
1/Sn -1/S(n-1)=2
所以{1/Sn}是等差数列.
S1=a1=1
1/Sn=1/a1+(n-1)d
1/Sn=1/1+(n-1)*2
1/Sn=2n-1
Sn=1/(2n-1)
S(n-1)=1/[2(n-1)-1]
S(n-1)=1/(2n-3)
an=Sn-S(n-1)
=1/(2n-1)-1/(2n-3)
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