已知{an}的前n项和为Sn,a1=1.且3an-1+2Sn=3求a1,a2的值,并求数列{an}的通项公式.
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已知{an}的前n项和为Sn,a1=1.且3an-1+2Sn=3求a1,a2的值,并求数列{an}的通项公式.
![已知{an}的前n项和为Sn,a1=1.且3an-1+2Sn=3求a1,a2的值,并求数列{an}的通项公式.](/uploads/image/z/3223753-25-3.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%3D1.%E4%B8%943an-1%2B2Sn%3D3%E6%B1%82a1%2Ca2%E7%9A%84%E5%80%BC%2C%E5%B9%B6%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.)
3a1+2S2=3a1+2(a2+a1)=3×1+2(a2+1)=3,a2=-1
由3an-1+2Sn=3得Sn=3/2-(3/2)a (n-1)
an=sn-s(n-1)=3/2-(3/2)a (n-1)-[3/2-(3/2)a (n-2)]=(3/2)×[a(n-2)-a(n-1)]
2an=3a(n-2)-3a(n-1),2an-6a(n-1)=3a(n-2)-9a(n-1) ,an-3a(n-1)=(-3/2)[a(n-1)-3a(n-2)]
数列{an-3a(n-1)}为等比数列,公比=-3/2,首项=a2-3a1=-1-3×1=-4,项数为n-2,
∴an-3a(n-1)=-4×(-3/2)^( n-2-1)= 4×(3/2)^( n-3),
数列{an-3a(n-1)}的和=-4×[1-(-3/2) ^(n-2)]/[1-(-3/2)]=2/5+2/5×(3/2)^( n-2)
又数列{an-3a(n-1)}的和=an-3a(n-1)+a(n-1)-3a(n-2)+a(n-2)-3a(n-3)+……+a2-3a1
=an-a1-2[a(n-1)+a(n-2)+a(n-3)+……+a2+a1]
=an-a1-2s(n-1)
=3an-a1-2sn=3an+3a(n-1)-a1-[3a(n-1)+2sn]=3an+3a(n-1)-1-3
=3[an+a(n-1)]-4
∴3[an+a(n-1)]-4=2/5+2/5×(3/2)^( n-2),
an+a(n-1)-22/15=(2/15)×(3/2)^( n-2)=1/5+(1/5)×(3/2)^( n-3)
an+a(n-1)-1/5=(1/5)×(3/2)^( n-3)
数列{an+a(n-1)-1/5}为等比数列,公比=3/2,首项=a2+a1-1/5=-1+1-1/5=-1/5,项数为n-2,
[an+a(n-1)-1/5]-[a(n-1)+a(n-2)-1/5]+[a(n-2)+a(n-3)-1/5]-[a(n-3)+a(n-4)-1/5]+……+[a3+a2-1/5]
-[a2+a1-1/5]=an-a1=an-1
=[an+a(n-1)-1/5]+[a(n-2)+a(n-3)-1/5]+……+[a3+a2-1/5] -[a(n-1)+a(n-2)-1/5]-……-[a2+a1-1/5]
=[an+a(n-1)-1/5]+[a(n-2)+a(n-3)-1/5]+……+[a3+a2-1/5]-[a(n-1)+a(n-2)-1/5]-……-[a2+a1-1/5]
=(1/5)×(3/2)^( n-3)+(1/5)×(3/2)^( n-5)+(1/5)×(3/2)^( n-7)+……+(1/5)×(3/2)^2+(1/5)×(3/2)^0
-(1/5)×(3/2)^( n-4)-(1/5)×(3/2)^( n-6)-(1/5)×(3/2)^( n-8)-……-(1/5)×(3/2)^3-(1/5)×(3/2)^1
-(1/5)×(3/2)^(-1)
上式中各加数和减数均为等比数列,公比=(3/2)^2=9/4,首项分别为(1/5)×(3/2)^0=1/5,
(1/5)×(3/2)^(-1)=(1/5)×(3/2)项数为(n-2)/2
因此,上式=an-1={(1/5)×[1-(9/4)^(n-2)/2]/[1-(3/2)]-{(1/5)×(3/2)[1-(9/4)^(n-2)/2]/[1-(3/2)]
=(1/5)[(9/4)^(n-2)/2-1]=(1/5)[(3/2)^(n-2)-1]=(1/5)(3/2)^(n-2)-1/5
所以,an=(1/5)(3/2)^(n-2)-1/5+a1=(1/5)(3/2)^(n-2)+4/5
an=sn-s(n-1)=(3/2)×[a(n-2)-a(n-1)]
由3an-1+2Sn=3得Sn=3/2-(3/2)a (n-1)
an=sn-s(n-1)=3/2-(3/2)a (n-1)-[3/2-(3/2)a (n-2)]=(3/2)×[a(n-2)-a(n-1)]
2an=3a(n-2)-3a(n-1),2an-6a(n-1)=3a(n-2)-9a(n-1) ,an-3a(n-1)=(-3/2)[a(n-1)-3a(n-2)]
数列{an-3a(n-1)}为等比数列,公比=-3/2,首项=a2-3a1=-1-3×1=-4,项数为n-2,
∴an-3a(n-1)=-4×(-3/2)^( n-2-1)= 4×(3/2)^( n-3),
数列{an-3a(n-1)}的和=-4×[1-(-3/2) ^(n-2)]/[1-(-3/2)]=2/5+2/5×(3/2)^( n-2)
又数列{an-3a(n-1)}的和=an-3a(n-1)+a(n-1)-3a(n-2)+a(n-2)-3a(n-3)+……+a2-3a1
=an-a1-2[a(n-1)+a(n-2)+a(n-3)+……+a2+a1]
=an-a1-2s(n-1)
=3an-a1-2sn=3an+3a(n-1)-a1-[3a(n-1)+2sn]=3an+3a(n-1)-1-3
=3[an+a(n-1)]-4
∴3[an+a(n-1)]-4=2/5+2/5×(3/2)^( n-2),
an+a(n-1)-22/15=(2/15)×(3/2)^( n-2)=1/5+(1/5)×(3/2)^( n-3)
an+a(n-1)-1/5=(1/5)×(3/2)^( n-3)
数列{an+a(n-1)-1/5}为等比数列,公比=3/2,首项=a2+a1-1/5=-1+1-1/5=-1/5,项数为n-2,
[an+a(n-1)-1/5]-[a(n-1)+a(n-2)-1/5]+[a(n-2)+a(n-3)-1/5]-[a(n-3)+a(n-4)-1/5]+……+[a3+a2-1/5]
-[a2+a1-1/5]=an-a1=an-1
=[an+a(n-1)-1/5]+[a(n-2)+a(n-3)-1/5]+……+[a3+a2-1/5] -[a(n-1)+a(n-2)-1/5]-……-[a2+a1-1/5]
=[an+a(n-1)-1/5]+[a(n-2)+a(n-3)-1/5]+……+[a3+a2-1/5]-[a(n-1)+a(n-2)-1/5]-……-[a2+a1-1/5]
=(1/5)×(3/2)^( n-3)+(1/5)×(3/2)^( n-5)+(1/5)×(3/2)^( n-7)+……+(1/5)×(3/2)^2+(1/5)×(3/2)^0
-(1/5)×(3/2)^( n-4)-(1/5)×(3/2)^( n-6)-(1/5)×(3/2)^( n-8)-……-(1/5)×(3/2)^3-(1/5)×(3/2)^1
-(1/5)×(3/2)^(-1)
上式中各加数和减数均为等比数列,公比=(3/2)^2=9/4,首项分别为(1/5)×(3/2)^0=1/5,
(1/5)×(3/2)^(-1)=(1/5)×(3/2)项数为(n-2)/2
因此,上式=an-1={(1/5)×[1-(9/4)^(n-2)/2]/[1-(3/2)]-{(1/5)×(3/2)[1-(9/4)^(n-2)/2]/[1-(3/2)]
=(1/5)[(9/4)^(n-2)/2-1]=(1/5)[(3/2)^(n-2)-1]=(1/5)(3/2)^(n-2)-1/5
所以,an=(1/5)(3/2)^(n-2)-1/5+a1=(1/5)(3/2)^(n-2)+4/5
an=sn-s(n-1)=(3/2)×[a(n-2)-a(n-1)]
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