an=SnSn-1 a1=2 9
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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
(1)易得Sn不等于0,由Sn-Sn-1=2SnSn-1,都除以SnSn-1,可以得到1/Sn-1/Sn-1=-2所以1/Sn为等差数列(2)1/S1=1/a1=1利用等差数列求通项的公式求出1/Sn
1.sn-s(n-1)+2SnSn-1=01/sn-1/s(n-1)=2所以1/sn是以1/s1=2为首项公差为2的等差数列即sn=1/2nan=-1/n(2n-2)(n≥2)bn=1/nbn^2=1
∵Sn-Sn-1=-anSn-Sn-1=-an/2∴d=1/Sn-1/Sn-1=(Sn-Sn-1)/SnSn-1=21/S1=1/a1=2∴{1/Sn}为首项=2,公差=2的等差数列
因为An=Sn-Sn-1.所以Sn-Sn-1+Sn*Sn-1=0,等式两边同时除以Sn*Sn-1得:1/Sn-1/Sn-1+=1,所以1/Sn为等差数列.因为a1=1/2.所以S1=1/2,1/S1=
an+2Sn·S(n-1)=0(n≥2),Sn-S(n-1)=an所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0即1/s
(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)∵Sn-Sn-1=2SnSn-1∴1Sn−1−1Sn=2即1Sn−1Sn−1=−2(常数)∴{1Sn}为等差数列  
当n=1时,an=1/2;当n>=2时,an=-1/(2n*(n-1));sn=1/2n;可以用归纳假设证明,由于没悬赏分,所以就不给你写详细过程了,哈哈哈.按着这个思路自己好好写下去,相信你没问题的
an=-2sn(sn-an)则2sn2-2snan+an=0解得:an=(5-3n)/4sn=(7n-3n2)/8
an+2SnS(n-1)=0Sn-S(n-1)=-2SnS(n-1)1/Sn-1/S(n-1)=21/Sn-1/S1=2(n-1)1/Sn=2nSn=1/(2n)Sn.S(n+1)=1/(2n).1/
an+2Sn·S(n-1)=0(n≥2)Sn-S(n-1)=an所以Sn-S(n-1)+2Sn·S(n-1)=0(n≥2)两边同时除以Sn·S(n-1),得1/S(n-1)-1/sn+2=0即1/Sn
∵An+2SnS(n-1)=0(n≥2)∴Sn-S(n-1)+2SnS(n-1)=0(n≥2)∴S(n-1)=Sn+2SnS(n-1)(n≥2)两边同时除以SnS(n-1),S(n-1)/[SnS(n
(Ⅰ)数列{1Sn}是以2为首项,2为公差的等差数列.证明如下:∵n≥2时,an+2SnSn-1=0,∴Sn-Sn-1+2SnSn-1=0∴1Sn-1Sn−1=2∵a1=12,∴1S1=2∴数列{1S
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/a1=
an=Sn-Sn-1代入然后同时除以SnSn-1即可可以证明是等差数列求an利用一问的结论求出1/an然后倒数就是an了数列主要利用前N项和做差公式牢记
liman/(Sn^2)=lim(Sn^2-1)/(Sn^2)=1-lim1/(Sn^2)lim(Sn^2)=无穷,所以最后=1
那个等式是不是Sn-S(n-1)+2Sn*S(n-1)=0啊(1)两边同除以Sn*S(n-1)得1/Sn-1/S(n-1)=2,由此即得{1/Sn}是等差数列,(2)由(1)可得1/Sn=2n,所以S
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-
1an=-SnSn-1所以Sn-Sn-1=-SnSn-11(/Sn-1)-1/Sn=-11/Sn-1(/Sn-1)=1所以{1/Sn}是等差数列2s1=a1=1/2所以1/Sn=1/2+(n-1)=n