已知数列sn是公差为2的等差数列,且a1 1.a3 1,a7 1成等比数列
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![已知数列sn是公差为2的等差数列,且a1 1.a3 1,a7 1成等比数列](/uploads/image/f/4267503-63-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97sn%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BA2%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E4%B8%94a1+1.a3+1%2Ca7+1%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97)
数列an满足条件:A1=1,A2=r(r>0)数列{an+an+1}是公差为d的等差数,令bn=an+an+1即首项b1=a1+a2=1+rb3=a3+a4=b1+2d=1+r+2db5=a5+a6=
1、√S1=√a1√S2=√(a1+a2)=√a1+2(1)√S3=√(a1+a2+a3)=√(3a2)=√a1+4(2)由(1)得a1+a2=a1+4√a1+4√a1=(a2-4)/4代入(2)√(
结果是an=4(2n+1);首先由s1,s2,s3的关系可列出两个方程,关于a1,a2,a3.和已知的2a2=a1+a3联立,求出a1=4.接下来,利用根号sn是等差数列,推导出s(n)和a1的关系,
1、由题意,得(a1+2)/2=√(2a1)整理,得(a1-2)²=0a1-2=0a1=2(an+2)/2=√(2Sn)整理,得8Sn=(an+2)²8Sn-1=[a(n-1)+2
因为an=Sn-S(n-1)又因为an=[√Sn+√S(n-1)]/2所以Sn-S(n-1)=[√Sn+√S(n-1)]/2==>[√Sn-√S(n-1)][√Sn+√S(n-1)=[√Sn+√S(n
a1=1,a2=3/2,a3=7/4an=2-(1/2)^(n-1)Tn=n/(2^(n-1))+(1/2)^(n-2)+n^2+n-4
再问:公差4n吧?再问:-4n再答:怎么会呢,比如N=2是不是比N=1差-4公差是一个不变的数,4N中N是可变的嘛
a(n+1)-an=22^a(n-1)÷2^an=2^[a(n+1)-an]=2²=4所以是等比数列,q=42^a1=2所以Sn=2*(1-4^n)/(1-4)=2(4^n-1)/3
(1)∵数列a[n]的前n项和为S[n],且满足a[n]+2S[n]S[n-1]=0,n≥2∴S[n]-S[n-1]+2S[n]S[n-1]=0两边除以S[n]S[n-1],得:1/S[n-1]-1/
sn=2n^2-n,bn=sn/(n+p)=(2n^2-n)/(n+p)b1=1/(1+p),b2=6/(2+p),b3=15/(3+p).bn是等差数列,则b1+b3=2b2,即1/(1+p)+15
1)由题意得,a1=1,当n>1时,sn=an^2/2+an/2sn-1=a(n-1)^2/2+a(n-1)/2,∴sn-sn-1=an^2/2-a(n-1)^2/2+an/2-a(n-1)/2即(a
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
(1)an是Sn与2的等差中项即a1=2sn=2an-2所以s(n-1)=2a(n-1)-2an=sn-s(n-1)=2a(n-1)所以an为等比数列公比为2首项为2则an=2^n而点P(bn,bn+
(1)2an=n+Sn2a(n+1)=n+1+S(n+1)相减得2【a(n+1)-an】=1+a(n+1)a(n+1)=2an+1b(n+1)=a(n+1)+1=2(an+1)=2bna1=1an=2
因:Sn是An和1的等差中项所以有:2Sn=An+1即:Sn=(An+1)/2An=Sn-S(n-1)=(An+1)/2-[A(n-1)+1]/2=[An-A(n-1)]/2An=-A(n-1)A1=
证:设等比数列{an}公比为q,对于数列{bn},对数有意义,q>0an=a1×q^(n-1)n=1时,b1=log3(a1)=log3(81)=4n≥2时,bn=log3(an)=log3(a1×q
a1+a2+...+an=(1/2)(an²+an)a1+a2+...+a(n-1)=(1/2)(a(n-1)²+a(n-1))两式相减得an=(1/2)(an²+an)
数列{an+Sn}是公差为2的等差数列∴an+Sn=a1+s1+2(n-1)=1+1+2n-2=2n∵当n=2时,a2+a2+a1=4∴a2=3/2(2)当n>=2时由an+Sn=2n得a(n-1)+
1数列{an+sn}是公差为2的等差数列,数列{an+sn}首项a1+s1=2a1=2,数列{an+sn}的通项=2+2(n-1)=2n,a2+s2=a2+a2+a1=4,a2=3/2,a3+s3=a