已知a1=1 3,Sn=n(2n-1)·an,求an.Sn
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证明,因为A(n+1)=(n+2)/n*Sn所以Sn=n*A(n+1)/(n+2)S(n-1)=(n-1)*An/(n+1)所以An=Sn-S(n-1)=n/(n+2)*A(n+1)-(n-1)/(n
(1)a2=4,方法就是取n=2,S2=a1+a2来算(2)2Sn=na(n+1)-n^3/3-n^2-2n/32an=Sn-S(n-1)an=n*a(n+1)/n+1-nan/n=a(n+1)/n+
(Ⅰ)由Sn+1=2Sn+n+5(n∈N*)得 Sn=2Sn-1+(n-1)+5(n∈N*,n≥2)两式相减得 an+1=2an+1,∴an+1+1=2(an+1)即 &
(Ⅰ)令n=1,得2a1-a1=a12,即a1=a12,∵a1≠0,∴a1=1,令n=2,得2a2-1=1•(1+a2),解得a2=2,当n≥2时,由2an-1=Sn得,2an-1-1=Sn-1,两式
(I)由已知Sn+1=2Sn+n+5(n∈N*),可得n≥2,Sn=2Sn-1+n+4两式相减得Sn+1-Sn=2(Sn-Sn-1)+1即an+1=2an+1从而an+1+1=2(an+1)当n=1时
因为Sn=n^2*an.1Sn-1=(n-1)^2*an-1n≥2.21-2:an=n^2*an-(n-1)^2*an-1(n^2-1)*an=(n-1)^2*an-1(n+1)*an=(n-1)*a
(1)∵Sn+1=2Sn+3n+1,∴当n≥2时,Sn=2Sn-1+3(n-1)+1,两式相减得an+1=2an+3,从而bn+1=an+1+3=2(an+3)=2bn(n≥2),∵S2=2S1+3+
评析:本页那位热心网友写错了:在得出an+1=3(a(n-1)+1)后,应将a2=8带入求值,因为前面a(n-1),n应大于等于二,所以a1不能算入通项公式中,应检验是否符合n大于等于二时的通项公式,
an=-Sn.S(n-1)Sn-S(n-1)=-Sn.S(n-1)1/Sn-1/S(n-1)=11/Sn-1/S1=n-11/Sn=nSn=1/n
a1+2a2+3a3+…+nan=(n-1)Sn+2n,①n=1时a1=2,n>1时a1+2a2+3a3+…+(n-1)a=(n-2)S+2(n-1),②①-②,nan=(n-1)Sn-(n-2)S+
sn=n^2ans(n-1)=(n-1)^2*a(n-1)sn-s(n-1)=n^2an-(n-1)^2*a(n-1)=an(n^2-1)an=(n-1)^2a(n-1)(n+1)an=(n-1)a(
为了避免混淆,我把下角标放在内.首先从数列本身的基本意义出发a=S-S其次,从已知a=S(n+2)/n出发a=S*(n+1)/(n-1)因此S-S=S*(n+1)/(n-1)移项整理S=S
由题意,S(n)-S(n-1)=2a(n+1)-2a(n),即a(n)=2a(n+1)-2a(n),于是a(n+1)=a(n)*3/2,即a(n)是公比是q=3/2的等比数列,且首项是a(1)=1,所
2an=1/n(n+1)=1/n-1/(n+1)2S=2a1+2a2+2a3+.+2an=(1/1-1/2)+(1/2-1/3)+(1/3-1/4)...+(1/n-1/(n+1))=1-1/(n+1
(1)S1=a1=-23,∵Sn+1Sn=an-2(n≥2,n∈N),令n=2可得,S2+1S2=a2-2=S2-a1-2,∴1S2=23-2,∴S2=-34.同理可求得S3=-45,S4=-56.(
Sn-1=(n-1)(n-1)an-1Sn-Sn-1=an=nnan-(n-1)(n-1)an-1(nn-1)an=(n-1)(n-1)an-1an=(n-1)/(n+1)*(n-2)/(n-1)*…
(1)S1=a1=(2a1/a1)-1=1S2=2a2/a1-1=2a2-1=a1+a2=1+a2所以2a2-1=1+a2a2=2(2)Sn=(2an/a1)-1=2an-1Sn-1=(2an-1/a
Sn-S(n-1)=nan-(n-1)a(n-1)-4n+4=an(n-1)an-(n-1)a(n-1)=4(n-1)an-a(n-1)=4所以an=4n-3
an=sn-s(n-1)代入得Sn=2S(n-1)+2^n,即Sn/2^n=S(n-1)/2^(n-1)+1所以Sn=(n+1/2)*2^n,所以an=Sn-S(n-1)=n*2^n+2^(n-1).
n≥2时,SnS(n-1)+(1/2)an=02SnS(n-1)+Sn-S(n-1)=0等式两边同除以SnS(n-1)2+1/S(n-1)-1/Sn=01/Sn-1/S(n-1)=2,为定值1/S1=