若等差数列an前n项和sn满足s412 s9 36
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没有分不给你解就提示你下吧3+4=2+5
(1)An=a1+(n-1)d由a3=7,可得a1+2d=7由a5+a7=26,可得a1+5d=13,即a1=3,d=2,则An=2n+1Sn=na1+n(n-1)d/2即Sn=n^2+2n
因为数列{an}为等差数列,且a1=1,则由等差数列性质可得:前n项和Sn=a1n-(n(n-1)/2)*D即Sn=n-(n(n-1)/2)*D,S2n=2n-(2n(2n-1)/2)*D且S2n/S
由lg(Sn+1)=n可得:Sn=10^n-1.n=1时,a1=S1=9,n≥2时,an=Sn-S(n-1)=10^n-1-(10^(n-1)-1)=9×10^(n-1)所以an=9×10^(n-1)
1、an=-2n+21;n=10时,Sn=100解析:因a4+a6=22,故a5=11,d=(11-15)/2=-2,Sn=(20-n)n2、bn=20x3*(n-1);Tn=10x(3*n-1)解析
答案为ASn=((a1+an)/2)*nan=a1+(n-1)d根据上式得出:Sn=(2a1+(n-1)d)*n/2=a1*n+n方*d/2-n*d/2limSn/n方=lim(2a1*n+n方*d-
http://zhidao.baidu.com/question/88231937.html?fr=qrl&cid=983&index=2S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=
S1=a1=-(2/3),S2+1/S2+2=a2,因为S2=(a1+a2),所以S2+1/S2+2=S2-a1=S2+2/3,解得S2=-(3/4),同理,S3+1/S3+2=a3=S3-S2=S3
a5+a7=a3+2d+a3+4d=2a3+6d2*7+6d=2614+6d=266d=12d=2a1=7-2*2=3an=3+2nsn=(a1+an)*n/2=(3+3+2n)*n/2=(6+2n)
等比数列{an}中,前n项和为sn,已知S1,S3,S2成等差数列,求{an}的公比Q.已知a1-a3=3,求sn?S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S1,S3,S2成等差数列
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
(1)设等差数列{an}的公差为d,∵前n项和Sn满足S3=0,S5=-5,∴3a1+3d=05a1+10d=−5,解得a1=1,d=-1.∴an=1-(n-1)=2-n.(2)1a2n−1a2n+1
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
唉,你太粗心了吧~我给你修正下(向我现在这样的好人不多了哈哈~!)Sm/Sn=(m^2)/(n^2),求am/an?对吧,很简单的呦am/an=2am/(2an)=a1+a2m-1/(a1+a2n-1
⑴∵等差数列{an}满足a3+a4=16,a4+a5=20,∴2a1+5q=162a1+7q=20∴a1=3,q=2∴Sn=na1+n(n-1)d/2=n²+2n⑵1/Sn=1/(n
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
由等差数列的通项公式可得a2+a5+a17+a22b8+b10+b12+b16=2(2a1+21d)2(2b1+21d′)=a1+a22b1+b22=22(a1+a22)222(b1+b22)2=S2
n=1时,a1=S1=a+bn≥2时,Sn=a×n²+bnS(n-1)=a×(n-1)²+b两式相减得:an=Sn-S(n-1)=2a×n-a∴a(n-1)=2a×(n-1)-a∴
由题意可得S14T14=14(a1+a14)214(b1+b14)2=2a72b7=a7b7=3×14+24×14−5=4451,故答案为:4451.