设数列an满足n 1=an^2-nan 1(1)an=2时,求a1,a2,a3

（Ⅰ）由已知，当n≥1时，an+1=[（an+1-an）+（an-an-1）+…+（a2-a1）]+a1=3（22n-1+22n-3+…+2）+2=22（n+1）-1．而a1=2，所以数列{an}的通

设bn=an/nSn=n^2-2n-2bn=sn-sn-1=2n-3b1=s1=-3所以an=n(2n-3)n>=2an=-3n=1

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={

a1×a2×a3×a4=a1+a2+a3+a41×1×2×a4=1+1+2+a4a4=4a2×a3×a4×a5=a2+a3+a4+a51×2×4×a5=1+2+4+a57a5=7a5=1=a1a3×a

题目不对吧.,(an+1)(an)=(an-1)(an-2+2),要是an=（an-2）+2那an+1=an-1了.还有,这种+1,+2的,到底是n+1,n+2,还是就是+1,+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)]

（Ⅰ）由题意可得数列{an}是首项为1，公比为3的等比数列，故可得an=1×3n-1=3n-1，由求和公式可得Sn=1×(1−3n)1−3=12(3n−1)；（Ⅱ）由题意可知b1=a2=3，b3=a1

(1)根据题意,有An=(An-An-1)+(An-1-An-2)+…+（A2-A1）+A1=3-2^(2n-3)+3-2^(2n-5)+…+(3-2^3)+2再用分组求和法：=3n-【2^(2n-3

（1）2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a

多写一项a1+2a2+2^2a3+...+2^n-2an-1=n-1/2,两式相减,有2^n-1an-2^n-2an-1=1/2,即2^nan-2^n-1an-1=1,所以2^nan=2a1+（n-1

an=lg5/√3^2n+1=lg5+(n+1/2)lg3a(n+1)=lg5+(n+1+1/2)lg3,a(n+1)-a(n)=lg3(常数）,an是等差数列.

由题意,显然该等比数列的公比不会是负数,也不会是小于一的数.前者不会满足等差数列要求,后者末项趋于零,不合理.故公比大于一,故等差数列是递增的即公差大于0.又a5*a5=a3*an1即36=a3*an

x=anf(x)=a(n+1)代入函数方程a(n+1)=an^2+2ana(n+1)+1=an^2+2an+1=(an+1)^2满足平方递推数列定义,因此数列{an+1}是平方递推数列.a1+1=10

由题意得：an-a(n-1)=3·2^(2n-3)a(n-1)-a(n-2)=3·2^(2n-5)..a2-a1=3·2^1叠加得：an-a1=3·[2^1+2^3+.+2^(2n-3)]注意：共n-

(1)由bn=√(4an+1)推出bn^2=4an+1即4an=bn^2-1则4a(n+1)=b(n+1)^2-1那么条件4a(n+1)=4an+2√(4an+1)+1就等价于b(n+1)^2-1=b

∵2nan+1=(n+1)an,∴a(n+1)/an=(n+1)/2n,∴a2/a1=2/2a3/a2=3/2×2a4/a3=4/2×3a5/a4=5/2×4……an/a(n-1)=n/2(n-1)两

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

1、a(n+1)/an=(n+2)/(n+1)a(n+1)/(n+2)=an/(n+1)设cn=an/(n+1)则c(n+1)=a(n+1)/(n+2),且c1=a1/(1+1)=1即c(n+1)=c

a(n+1)=an^2+6an+6=(an+3)^2-3,即a(n+1)+3=(an+3)^2,从而log5[a(n+1)+3]=2log5(an+3)而cn=log5(an+3),则结合上式即得c(

由递推式有a2-a1=3*2a3-a2=3*2*4a4-a3=3*2*4^2.an-a(n-1)=3*2*4^(n-2)累加得an-a1=2*4^(n-1)-8得an=2*4^(n-1)-6于是bn=