设bn等于绝对值求tn

Tn+Bn/2=1Tn=1-Bn/2T(n-1)=1-B(n-1)/2Tn-T(n-1)=Bn=-Bn/2+B(n-1)/22Bn=-Bn+B(n-1)3Bn=B(n-1)Bn/B(n-1)=1/3n

1.Sn=3/2bn-3/2S(n-1)=3/2b(n-1)-3/2bn=Sn-S(n-1)=3/2bn-3/2b(n-1)bn=3b(n-1)所以{bn}是等比数列,公比3而b1=S1=3/2b1-

S3=3a1+3*2*d/2=3a1+3ds5=5a1+5*4*d/2=5a1+10da3*b3=(a1+2d)/S3=(a1+2d)/(3a1+3d)S3+S5=3a1+3d+5a1+10d=8a1

楼主没有给an的表达式呀

n=b^2n,Tn=b^2+b^4+b^6+……+b^2n=b^2n(1-b^2n)/（1-b^2)所以1-bn=1-b^2n所以（1-bn）/Tn=(1-b^2n)/{b^2(1-b^2n)/(1-

a(n)*a(n+1)=(6n-5)(6n+1)1/[(6n-5)(6n+1)=(1/6)*[1/(6n-5)-1/(6n+1)]Tn=(1/6)*[1-1/7+1/7-1/13+1/13-1/19+

(1)bn=2n/(2n-1)T1=2,T2=8/3,T3=16/5(2)注意:2k/(2k-1)=1+1/(2k-1)>1+1/2k=(2k+1)/2k于是:Tn^2=(2n/(2n-1))(2n/

由Sn=n²+2n+1易得a1=4（当n=1）an=2n-1（当n≥2）所以b1=8（当n=1）bn=（2n-1）*2^nTn=8+3*2^2+5*2^3+7*2^4+...+(2n-1)*

an=Sn-S(n-1)=3an+2-3a(n-1)-2an=3/2a(n-1)a1=3a1+2a1=-1an=(-1)*(3/2)^(n-1)anbn=-n*(3/2)^(n-1)Tn=-1(3/2

当n≥2时,有bn=Tn-T(n-1)所以由6Tn=(3n+1)bn+2得6T(n-1)=(3(n-1)+1)b(n-1)+2上两式相减得6(Tn-T(n-1)=(3n+1)bn-(3n-2)b(n-

Tn=b1+b2+...+bn=(3/a1a2)+.+3/[ana(n+1)]=3[1/a1a2+1/a2a3+...+1/ana(n+1)]=3[1/(1*7)+1/(7*13)+...+1/(6n

n=1时,a1=3n>=2时Tn=n^2+n+1.(1)T(n-1)=(n-1)^2+(n-1)+1.(2)两式相减得an=2n(n>=2)n=1代入,a1=2,不符合综合得n=1,a1=3n>=2,

Tn=2+2*4+3*8+4*16+.+(n-1)*(2的n-1次方)+n*（2的n次方）2*Tn=4+2*8+3*16+.+（n-1）*（2的n次方）+n*（2的n+1次方）用二式减一式得Tn=n*

a(n)=aq^(n-1),a>0,q>0.a+aq=a(1)+a(2)=2[1/a(1)+1/a(2)]=2[1/a+1/(aq)]=2(q+1)/(aq),a=2/(aq),q=2/a^2,a(n

将an带入bn得bn=n/3*2^(n-1)；将Tn展开为Tn=1/3（1+2/2+3/2^2+4/2^3+...+n/2^(n-1)）---此为1式然后等是两边同时1/2*Tn=1/3（1/2+2/

Cn=an/bn=（4n-2)/[2/4^(n-1)]=(n-1)4^(n-1)Tn=0+1*4+2*4^2+3*4^3+.+(n-1)4^(n-1)4Tn=1*4^2+2*4^3+3*4^4……(n

因为点(bn,Tn)在直线y=-1/2x+1上所以Tn=-1/2bn+1即2Tn=-bn+2因为Tn-T(n-1)=bn所以2Tn=-Tn+T(n-1)+2即3Tn=T(n-1)+2等式两侧都减3即3

Tn=b1+b2+...+bn=(3/a1a2)+.+3/[ana(n+1)]=3[1/a1a2+1/a2a3+...+1/ana(n+1)]=3[1/(1*7)+1/(7*13)+...+1/(6n

a4+a6=2a5=16,a4a6=60,解得a4=6,a6=10,2d=a6-a4=4,d=2,an=a4+(n-4)x2=2n-2.sn=n(n-1)/2*2=n(n-1)1/sn=1/n(n-1

(1)an=sn-s(n-1)=n^2+2n-(n-1)^2-2(n-1)=2n+1(2)bn=(2n+1-1)/2=nTn=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-