设数列an满足a1 2a2 2^2a3

（Ⅰ）由题意知数列{an}是首项为1，公比为3的等比数列，其通项公式为an=3n-1；数列{bn}满足b1=S1=4，n≥2时，bn=Sn-Sn-1=2n+1．所以，数列{bn}的通项公式为bn＝4，

（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

Sn=(1/4)(an+1)^2S(n-1)=(1/4)[a(n-1)+1]^2相减且an=Sn-S(n-1),所以4an=(an+1)^2-[a(n-1)+1]^2[a(n-1)+1]^2=(an+

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={

an=1/3^n

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)]

(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

（1）当n=1时,a1>=3=1+2,an>=n+2成立；当n>1时,an=（an-1）^2-nan-1+1,令S=an-（n+2）=（an-1）^2-nan-1+1-（n+2）=（an-1）^2-(

令n=1时,a1=1*2*3=6；依题意：a1+2a2+3a3+.+nan=n(n+1)(n+2),a1+2a2+3a3+.+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)两式相减,得

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-

∵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

a1＋3a2＋3²a3＋…＋3^(n－1)an=n/3a1＋3a2＋3²a3＋…＋3^(n－2)a(n－1)=(n－1)/3=n/3-1/3（n≥2）两式相减得：3^(n-1)an

a(n)=aq^(n-1),b(n)=na(1)+(n-1)a(2)+...+2a(n-1)+a(n),m=b(1)=a(1)=a,3m/2=b(2)=2a(1)+a(2)=2m+a(2),a(2)=

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

an满足an满足a1+2a2+3a3+...+nan=2^n所以有a1+2a2+3a3+...+（n-1）a（n-1）=2^（n-1）上面两式作减法有nan=2^n-2^(n-1)=2^(n-1)an

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=