在数列an中a1等于1前n项和为sn和an

Sn^2=an×(Sn-1/2)=(Sn-Sn-1)×(Sn-1/2)整理,得Sn-1-Sn=2SnSn-1等式两边同除以SnSn-11/Sn-1/Sn-1=2,为定值.1/S1=1/a1=1/1=1

（1）数列{an}中，a1=1，前n项和Sn＝n+23an，可知S2＝43a2，得3（a1+a2）=4a2，解得a2=3a1=3，由S3＝53a3，得3（a1+a2+a3）=5a3，解得a3=32(a

设：(An+1)+p(n+1)+q=4[An+pn+q]解得p=-1,q=0即An+1=4An-3n+1等价于(An+1)-（n+1）=4（An-n）若设Bn=An-n则Bn+1=4Bn则Bn=B1*

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

设{an}的公比为q,则a2=2q,a3=2q^2则(a2+1)^2=(a1+1)(a3+1)即(2q+1)^2=3(2q^2+1)解得q=1所以{an}为常数数列Sn=na1=2n

n>=2时,a(n+1)=3Sn(1),an=3S(n-1)(2)(1)-(2)：a(n+1)-an=3an,a(n+1)=4an所以,a2,a3,a4,…,an是公比为4的等比数列.

因数列{an}为等比，则an=2qn-1，因数列{an+1}也是等比数列，则（an+1+1）2=（an+1）（an+2+1）∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a

1/a(n+1)=an+2/2an=1/2+1/an所以,{1/an}是公差为1/2的等差数列1/an=1/a1+(n-1)*1/2=(n+1)/2an=2/(n+1)a(n+1)=2/(n+3)an

a2=2aa3=4a+1a4=6a+2.an=（2a+1）n-2（a-2）

在数列{an}中,a1=1,a(n+1)=2an+2^n,求数列的前n项和a(n+1)=2an+2^n同除以2^na(n+1)/2^n=2an/2^n+1a(n+1)/2^n-an/2^(n-1)=1

证：a(n+1)=2an/(an+1)1/a(n+1)=(an+1)/(2an)=(1/2)(1/an)+1/21/a(n+1)-1=(1/2)(1/an)-1/2=(1/2)(1/an-1)[1/a

1.变形即为a(n+1)-(n+1)=4(an-n),所以(an-n)是首项为1,公比为4的等比数列.2.令an-n=bn,则Sbn=（4^n-1）/(4-1),即San-1-2-…-n=（4^n-1

Sn=n^2/(3n+2)Sn-1=(n-1)^2/(3n-1)an=Sn-Sn-1=(3·n^2+n-2)/(9·n^2+3n-2)所以,当n接近正无穷时liman=1/3

将a[n+1]=S[n+1]-S[n]代人得到：S[n]=4(S[n+1]-S[n])+14S[n+1]=5S[n]-14(S[n+1]-1)=5(S[n]-1)(S[n+1]-1)/(S[n]-1)

因数列{an}为等比，则an=3qn-1，因数列{an+1}也是等比数列，则（an+1+1）2=（an+1）（an+2+1）∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a

设等比数列{an}的公比为q，则可得an=2•qn-1，故an+1=2•qn-1+1，可得a1+1=3，a2+1=2q+1，a3+1=2q2+1，由于数列{an+1}也是等比数列，故（2q+1）2=3

an=2S(n-1)an+S(n-1)=3S(n-1)Sn=3S(n-1)Sn/S(n-1)=3Sn是等比数列Sn/S(n-1)=3Sn/S1=3^(n-1)Sn=3^(n-1)(1)an=2S(n-

(a2+1)²=(a1+1)(a3+1)a1=2,设an公比q(2q+1)²=3(2q²+1)4q²+4q+1=6q²+32q²-4q+2=

设公比为q,a2²=a1*a3(a2+1)²=(a1+1)(a3+1)因为a1=2所以a2²=2a3(a2+1)²=3(a3+1)解得a2=2a3=2所以sn=

an+a(n+1)=6/5^(n+1)=(5+1)/5^(n+1)=1/5^n+1/5^(n+1)a(n+1)-1/5^(n+1)=-(an-1/5^n)a1-1/5^1=1/5-1/5=0an-1/