数列an满足an大于0 sn=1 2(an 1 an)
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![数列an满足an大于0 sn=1 2(an 1 an)](/uploads/image/f/5034477-21-7.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3an%E5%A4%A7%E4%BA%8E0+sn%3D1+2%28an+1+an%29)
an+2SnSn-1=0Sn-Sn-1+2SnSn-1=01/Sn-1-1/Sn+2=01/Sn-1/Sn-1=2{1/Sn}是以首项为1/a1=2公差为2的等差数列1/Sn=2+(n-1)*2=2n
An=Sn-Sn-1所以原式=Sn-Sn-1+2SnSn-1=0同时除以2SnSn-11/Sn-1/Sn-1=2所以1/Sn为等差数列1/S1=21/Sn=2+(n-1)*2=2n所以Sn=1/2n再
(1)因为2an=Sn*S(n-1)所以2(Sn-S(n-1))=Sn*S(n-1)两边同除Sn*S(n-1)整理的1/Sn-1/S(n-1)=-1/2(n>1)所以数列{1/Sn}是以1/Sn=1/
(1)当n=1时,S2=2µ*S1+1=2µ*a1+1,S2=2µ+1当n=2时,S3=2µ*S2+1,则S2+a3=2µ*S2+12&mi
an+2Sn*S(n-1)=0而an=Sn-S(n-1)∴Sn-S(n-1)+2Sn*S(n-1)=0同除以Sn*S(n-1)整理:1/Sn-1/S(n-1)=2∴{1/Sn}为等差数列,公差2,首项
(1)2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a
2Sn(Sn-An)=-An2SnSn-1=Sn-1-Sn1/Sn-1/Sn-1=2{1/Sn}便是一个等差数列,其首项为1/S1=1/A1=1/2得出的结果便是:Sn=2/(4n-3)An=2/(4
an=Sn-Sn-1=-SnS(n-1)(Sn-Sn-1)/[SnS(n-1)]=-11/S(n-1)-1/Sn=-11/Sn-1/S(n-1)=1,为定值.1/S1=1/a1=1/(1/2)=2数列
Sn²=an(Sn-1/2)an=Sn-Sn-1Sn²=(Sn-Sn-1)(Sn-1/2)=Sn²-SnSn-1-1/2*Sn+1/2*Sn-1SnSn-1=-1/2(S
An=Sn-S(n-1)=2An+(-1)^n-2A(n-1)+(-1)^(n-1)=2An-2A(n-1)得An=2A(n-1)根据此式知道An为等比数列公比为2求第一项S1=2A1-1=A1得A1
因为an,Sn,Sn-1/2成等比数列所以an*Sn-1/2=Sn^2因为an=Sn-Sn-1所以(Sn-Sn-1)*Sn-1=2Sn^2等式两边同除以Sn^2得[1-(Sn-1/Sn)](Sn-1/
An+2Sn*Sn-1=0Sn-Sn-1+2Sn*Sn-1=01/Sn-1-1/Sn+2=01/Sn=2nSn=1/2n(n>=2)An=1/(2n-2n^2)(n>=2)=1/2(n=1)
已知数列a‹n›首相a₁=3,通项a‹n›和前n项和S‹n›之间满足2a‹n›=S̸
由题意知:2an/[anSn-(Sn)²]=1(n>1)则:(Sn)²-anSn+2an=0(n>1)又因为:an=Sn-S(n-1)(n>1)所以:(Sn)²-[Sn-
an+2SnSn-1=0Sn-Sn-1+2SnSn-1=01/Sn-1/Sn-1=21/Sn=2+2(n-1)Sn=1/nan=Sn-Sn-1=1/n-1/(n-1)1/2n=1an=-1/[n(n-
∵s[n]=n^2a[n]∴s[n+1]=(n+1)^2a[n+1]将上述两式相减,得:a[n+1]=(n+1)^2a[n+1]-n^2a[n](n^2+2n)a[n+1]=n^2a[n]即:a[n+
/>n≥2时,Sn=n²×anS(n-1)=(n-1)²×a(n-1)an=Sn-S(n-1)=n²×an-(n-1)²×a(n-1)(n²-1)an