已知Sn=1^2-2^2 3^2
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∵Sn-Sn-1=√Sn+√Sn-1∴(√Sn)²-(√Sn-1)²=√Sn+√Sn-1(√Sn-√Sn-1)(√Sn+√Sn-1)=√Sn+√Sn-1∴√Sn-√Sn-1=1(n
因为但看1+2+3...+n这个数列,通项公式为n(n+1)/2=n^/2+n/2所以1=1/2(1^+1)1+2=1/2(2^+2)1+2+3=1/2(3^+3)以此类推,提出共因数1/2,合并括号
an+2Sn*Sn-1=0其中an=Sn-Sn-1代入上式:Sn-Sn-1+2Sn*Sn-1=0a1=1/2,故Sn和Sn-1≠0,上式两边同除以Sn*Sn-1得:1/Sn-1-1/Sn+2=0即:1
由题得:an=1/2(1/n-1/(n+1);所以:a1=1/2(1-1/2);a2=1/2(1/2-1/3);a3=1/2(1/3-1/4);.an=1/2(1/n-1/(n+1);sn=a1+a2
由Sn=Sn-1/2Sn-1+1,两边同时取倒数可得1/Sn=(2Sn-1+1)/Sn-11/Sn=2+1/Sn-1即1/Sn-1/Sn-1=2故{1/Sn}是首项为1/2,公差为2的等差数列1/Sn
第一个搞定我就不罗嗦了即1/Sn-1/Sn-1=2所以有1/Sn-1/Sn-1=21/Sn-1-1/Sn-2=21/Sn-2-1/Sn-3=2…………1/S2-1/S1=2叠加得1/Sn-1/S1=2
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
1.“满足(Sn-Sn+1)/Sn-1-Sn=2+1/an”根据这个式子,能化简成An+1/An=2An+1(注意这里及以后的An+1就是下标的意思)再进一步化简,能得到:An+1=2an+1再凑配能
Tn=n/(n+2)+(n+2)/n-2=4/n(n+2)=2[1/n-1/(n+2)]于是T1+T2+T3+……Tn=2[1-1/3+1/2-1/4+1/3-1/5+……+1/(n-1)-1/(n+
an=(2n+1)*3^na1=3*3^1a2=5*3^2a3=7*3^3.an=(2n+1)*3^nSn=3*3^1+5*3^2+7*3^3+.(2n+1)*3^n3Sn=3*3^2+5*3^3+7
我会我会Sn+1=Sn-2nSn+1Sn两边同除以Sn+1*Sn得1/Sn+1-1/Sn=2n以此类推1/Sn-1/Sn-1=2(n-1)1/Sn-1-1/Sn-2=2(n-2)...1/S2-1/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
(1)由sn=sn-12sn-1+1(n≥2),a1=2,两边取倒数得1Sn=1Sn-1+2,即1Sn-1Sn-1=2.∴{1sn}是首项为1S1=1a1=12,2为公差的等差数列;(2)由(1)可得
n>=2时:∵an=2Sn^2/[(2Sn)-1]∴Sn-(Sn-1)=2Sn^2/[(2Sn)-1]两边同时乘以(2Sn)-1并化简得2Sn(Sn-1)+Sn-(Sn-1)=0两边同时除以Sn(Sn
(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)*…
Sn=Sn-1/(2Sn-1+1)S1=A1=1S2=1/(2*1+1)=1/3S3=1/3/(2*1/2+1)=1/5假设Sk=1/(2k-1)Sk+1=1/(2k-1)/(2*1/2k-1+1)=
2(Sn+1)(Sn)/(Sn-Sn+1)=1上下除以(Sn+1)(Sn)得到2/(1/Sn+1-1/Sn)=11/(Sn+1)-1/Sn=2因此1/Sn+1为等差数列,1/S1=1/a1=1/21/