求数列{an}1 1² 2,1 2² 4

a(n+1)-an=2^n则an-a(n-1)=2^(n-1)……a2-a1=2^1相加an-a1=2^1+……+2^(n-1)=2*[1-2^(n-1)]/(1-2)=2^n-2a1=1所以an=2

1/a(n+1)=(an+2)/2an=1/2+1/an1/a(n+1)-1/an=1/2所以1/an是等差数列,d=1/21/an=1/a1+1/2*(n-1)=(n+1)/2an=2/(n+1)

a(n+1)=a(n)+2说明这是一个等差数列首项a(1)=-11,公差为2a(n)=a(1)+(n-1)×2=-11+2(n-1)=2n-13所以Sn=[a(1)+a(n)]×n/2=(n-12)n

Sn=1/2(an+1/an),an>0令x=1得：S1=1/2(a1+1/a1)解得a1=1注意到an=Sn-S(n-1),上式可化为：Sn=1/2(Sn-S(n-1)+1/(Sn-S(n-1)))

a1=2a1+a2+a3=12a2=4d=2an=2nbn=3^an=3^2n=9^n数列bn是以9为首项,公比=9的等比数列Sn=9(1-9^n)/(1-9)=(9^[n+1]-9)/8

a1=2,a1+a2+a3=12a2=4d=2an=2n2.Sn=2*3+4*3^2+6*3^3+……+2n*3^n3Sn=2*3^2+4*3^3+……+(2n-2)*3^n+2n*3^[n+1]相减

由于Sn=2^n则:S1=a1=2^1=2当n>=2时,an=Sn-S(n-1)=2^n-2^(n-1)=[2*2^(n-1)]-2^(n-1)=2^(n-1)又a1=2则:an=2^(n-1)(n>

由题意,Sn=n^2,则a1=1,S(n-1)=(n-1)^2=n^2-2n+1,n>=2an=Sn-S(n-1)=n^2-n^2+2n-1=2n-1,n>=2由于当n=1时,2n-1=1=a1所以,

a(n+1)=3an+1a(n+1)+1/2=3an+3/2=3(an+1/2)[a(n+1)+1/2]/(an+1/2)=3,为定值.a1+1/2=1/2+1/2=1数列{an+1/2}是以1为首项

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+1=an+2,则数列an是公差为2的等差数列a10=a1+9d=-11+9*2=7s10=（a1+a10）*10/2=-20

(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an-2n+11>0,解得：n

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=

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+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