an² 2an=4sn 3 求数列bn的前n项和

1)a1=34Sn=an^2+2an-34S(n-1)=a(n-1)^2+2a(n-1)-34Sn-4S(n-1)=an^2-a(n-1)^2+2an-2a(n-1)4an=an^2-a(n-1)^2

记Sn=a1+a2/2+a3/3+a4/4……+an/n=An+B,则a1=S1=A+B,当n>=2时,an/n=Sn-S(下标n-1)=An+B-[A(n-1)+B]=A,an=An,所以,an={

a(n+1)/an=5^nan=a1*(a2/a1)(a3/a2)(a4/a3).(an/an-1)=4*5¹5²5³.*5^(n-1)=4*5^[1+2+3+……(n-

解题思路：利用an=Sn-Sn-1来解答。解题过程：最终答案：略

an=nba(n-1)/(a(n-1)+n-1)an.a(n-1)+(n-1)an=nba(n-1)1+(n-1)[1/a(n-1)]=nb(1/an)(n-1)(1/a(n-1)+[1/(1-b)]

2an=an+1+an-1->{an}是等差数列a1=1/4,a2=3/4->an=(2n-1)/43an-an-1=(6n-3)/4-(2n-3)/4=n=3bn-bn-1->3(bn-an)=bn

a1=1an=an-1+3n-2an-1=an-2+3(n-1)-2...a2=a1+3*2-2左右分别相加an=a1+3*(n+n-1+...+2)-2*(n-1)an=1+3*(n+2)*(n-1

(an+2)/2=√(2Sn)两边平方整理：(an+2)²=8snn-1代换n(a(n-1)+2)²=8s(n-1)两式对应相减(an+2)²-(a(n-1)+2)

令f(x)=(x+4)/(2x-1)=x,解得：x1=-1,x2=2取F(x)=(x+1)/(x-2)则：F^-1(x)=(2x+1)/(x-1),那么g(x)=F.f.F^-1=(x+1)/(x-2

an=nba(n-1)/(a(n-1)+n-1)an.a(n-1)+(n-1)an=nba(n-1)1+(n-1)[1/a(n-1)]=nb(1/an)(n-1)(1/a(n-1)+[1/(1-b)]

a(n+1)-2an=3.5^n,则a2-2a1=3.5^1a3-2a2=3.5^2.a(n+1)-2an=3.5^n以上式子相加,得a(n+1)-a1-Sn=3.5+3.5^2+...+3.5^n=

稍等,题目不太清楚,能把数列的下标用括号括起来吗,这样容易弄混.再答：an=nba(n-1)/[a(n-1)+(n-1)]ana(n-1)=nba(n-1)-(n-1)an∵an≠0∴上式等号两边同时

待定系数法因为a（n+1）=2an-n^2+3n设a(n+1)+p(n+1)^2+q(n+1)=2(an+pn^2+qn)展开整理得a(n+1)=2an+pn^2+(q-2p)-(p+q)与原式一一对

a(n+2)+2an=3a(n+1)a(n+2)-a(n+1)=2a(n+1)-2an[a(n+2)-a(n+1)]/[a(n+1)-2an]=2∴数列{an+1-an}是等比数列a(n+1)-an=

A(n+1)=An+2(n+1)A(n+1)-An=2(n+1)即An-A(n-1)=2nA(n-1)-A(n-2)=2(n-1).A3-A2=2*3A2-A1=2*2以上各式相加得：An-A1=2*

（1）a(n+1)=3an/(2an+3)a1=1a2=3a1/(2a1+3)=3/5a3=3a2/(2a2+3)=3/7a4=3a3/(2a3+3)=3/9=1/3a5=3a4/(2a4+3)=3/

我表示一楼很挫,楼主既然问这个问题不是找你要答案你总得写点过程吧an+1=an^2两边同时取对数lgan+1=2lgan则lgan为等比数列lgan=lga1*2^(n-1)an=a1^(2^(n-1

a1=1a2=(3+2)/(1+4)=1……an=1则bn=2