设数列an是首项为1的正项数列,且当n 2

S[1]=a[1]=1/2(a[1]+1/a[1]),于是：a[1]=1=√1-√0S[2]=a[2]+1=1/2(a[2]+1/a[2]),于是：a[2]=√2-1,S[2]=√2S[3]=a[3]

（1）设数列{an}的公差为d，数列{bn}的公比为q，则由题意知a1b1=1(a1+d)(b1q) =4(a1+2d)(b1q2) =12 ，因为数列{an}各项为正数

设an=a1*q^n-1则lgan-1+lgan+1=lga1*q^n-2+lga1*qn=lga1^2*q2n-22lgan=2lga1*qn-1=lg(a1*qn-1)^2=lga1^2*q2n-

0=(n+1)[A(n+1)]^2-n[A(n)]^2+A(n+1)A(n)=n{[A(n+1)]^2-[A(n)]^2}+A(n+1)[A(n+1)+A(n)]=[A(n+1)+A(n)]{A(n+

（n+1）[a(n+1）]^2-n(an)^2+a(n+1)an=0＝＞〔(n+1)a(n+1)-n*an〕*[a(n+1)+an]=0(分解因式)因为{an}是首项为1的正项数列所以a(n+1)+a

∵a1=1∴a2=1,a3=1,a4=1,a5=1an=1

((n+1)An+1-nAn)(An+1+An)=0=>(n+1)An=nAn=A1=1所以An=1/n

n[a(n+1)]²+[a(n+1)]²-n(an)²+a(n+1)an=0n{[a(n+1)]²-(an)²}+[a(n+1)]²+a(n

(n+1)*a(n+1)^2-n*an^2+an*a(n+1)=0n*(a(n+1)^2-an^2)+a(n+1)^2+an*a(n+1)=0(a(n+1)+an)((n+1)*a(n+1)-n*an

(一）（1）由a1=1,S(n+1)=4an+2.可得：a1=1,a2=5,a3=16.a4=44.∴由bn=(an)/2^n得：b1=2/4,b2=5/4,b3=8/4,b4=11/4.显然,b1,

an=1000*(1/10)^(n-1)=10^3*10^(1-n)=10^(4-n)lgan=4-nbk=lga1+lga2+...+lgak=3+2+...+4-k=(3+4-k)*k/2=(7-

Sn=an-2;Sn-1=an-3;an=an-2-an-3;条件不足,a1,a2没有初始值吗

（1）由题意得Tn=1-an，①Tn+1=1-an+1，②∴由②÷①得an+1=1−an+11−an，∴an+1=12−an，∴1Tn+1-1Tn=11−an+1-11−an=11−12−an-11−

因为an,Sn,an^2成等差数列所以2Sn=an^2+an2an=2Sn-2S(n-1)=an^2+an-a(n-1)^2-a(n-1)得:(an-a(n-1))(an+a(n-1))-(an+a(

设公比为q,因为a1=1,即：a(n)=q^(n-1)则：S(n)=(1-q^n)/(1-q)若{Sn}为等差数列,设公差为d则：S(n)=S(n-1)+d即：d=S(n)-S(n-1)=(1-q^n

不是很清楚饿,你把基本的会了吧,~

（1）设数列{an}的公差为d，数列{bn}的公比为q，由题意得d+q＝12d+q2＝2，解得d＝1q＝0(舍) 或d＝−1q＝2，则an=1-n，bn=2n-1．（2）由（1）知，cn=a

Sn+an=-(1/2)n^2-(3/2)n+1n=1a1=-1/22Sn-S(n-1)=-(1/2)n^2-(3/2)n+12(Sn+(1/2)n^2+(1/2)n-1)=S(n-1)+(1/2)(

an=3+(n-1)da(n+1)=3+nd所以bn=6+(2n-1)d=(6-d)+2dn所以bn是等差数列b1=6-d+2d=6+d所以Sn=(b1+bn)n/2=(12+2dn)n/2=dn&s

根号Sn的通项公式是nSn=n^2an=Sn-Sn-1=n^2-(n-1)^2=2n-1