设等比数列的前n项积为Tn,若T4=1 2,T8=2,则T16=

sn=a1(1-q^n)/(1-q)tn=1/a1*[1-(1/q)^n]/(1-1/q)Sn/Tn=a1^2q^(n-1)

设{an}公差为d,{bn}公比为q,q>0a3+b3=a1+2d+b1q^2=1+2d+3q^2=17d=(16-3q^2)/2T3-S3=b1(1+q+q^2)-(a1+a1+d+a1+2d)=3

a3=a1*q^2,a6=a1*q^5,S3=a1(1-q^3)/(1-q),S6=a1(1-q^6)/(1-q),T3/T6=a3S6/a6S3=(1-q^6)/[(1-q^3)*q^3]=(1+q

a3=1+2db3=3q²所以1+2d+3q²=17T3=b1+b2+b3=3+3q+3q²S3=a1+(a1+d)+(a1+2d)=3+3d所以3+3q+3q²

An满足A2=2,A5=8a5-a2=3d=6d=2a1=a2-d=0an=2(n-1)b3=a3=4T3=7=b1+b2+b3b1+b2=3b2^2=b1b3=4b1=4(3-b2)b2=2所以公比

(1)S5=5a1+10d=5+10d=45,d=4,a3=1+2d=9.T3=b1+b2+b3=1+q+q^2=9-q,则q=-4或q=2.因为q>0,所以q=2.{an}的通项公式为：an=1+4

S5=(a1+a5).5/2=45,所以a5=17又因为a=a1+(5-1)d所以d=4,:所以an=a1+(n-1).d=4n-3T3=a3-b2=9-1.q=b1+b2+b3=1+q+q.q所以q

设等比数列{an}的公比为q侧：Sn=a1(q的n次方-1)/(q-1)Tn=1/a1+1/a2+,=1/a1[((1/q)的n次方-1)/(1/q-1)=[(q的n次方-1)/(q-1)]/[a1&

1)(a99-1)/(a100-1)

t(1)=a(1)=1-a(1),a(1)=1/2=t(1).t(n)=1-a(n)a(n)=1-t(n)a(n+1)=1-t(n+1)a(n+1)t(n)=[1-t(n+1)]t(n)=t(n+1)

设等差数列的等差为d,等比数列的等比是q则a3=b3a4-d=b4/q又∵a4=b4∴a4-d=a4/qa4-a4/q=d∵（S5-S3）/（T4-T2）=5∴(a5+a4)/(b4+b3)=(a4+

取N=13Sn=(a1+a13)X13/2Tn=(b1+b13)x13/2a1+a13=2Xa7同理b7则比值为92/79

设该数列为an首项为a1公比：q,则Sn=a1(1-q^n)/(1-q)倒数数列首项：1/a1,共比：1/q,Tn={1/a1[(1-(1/q)^n)]}/(1-1/q)=q(q^n-1)/[a1q^

∵a2014a2015-1＞0，∴a2014a2015＞1，又∵a2014−1a2015−1＜0，∴a2014＞1，且a2015＜1．T4028=a1•a2…a4028=（a1•a4028）（a2•a

（1）由题意得Tn=1-an，①Tn+1=1-an+1，②∴由②÷①得an+1=1−an+11−an，∴an+1=12−an，∴1Tn+1-1Tn=11−an+1-11−an=11−12−an-11−

a3a4a8=a1^3*q^(12)=(a1*q^4)^3T8=a1^8*q^(1+2+3+……+7)=a1^8*q^28T9=a1^9*q^(1+2+3+……+7+8)=a1^9*q^36=(a1*

Tn=1/a1+1/a2+……+1/anTn/q=1/a2+……+1/an+1/(q*an)Tn-Tn/q=1/a1-1/(q*an)Tn=q/a1(q-1)-1/an(q-1)

T5=a1*a2*a3*a4*a5=a1*a1q*a1q^2*a1q^3*a1q^4=(a1q^2)^5=32=2^5a1q^2=a3=2

Sn=a1*(1-(根号2)^n)/(1-根号2)Tn=（17Sn-S2n）/an+1将Sn=a1*(1-(根号2)^n)/(1-根号2)an+1=a1*根号2^n带入其中求解,得(17-17根号2^

T1=a1=1-a12a1=1a1=1/2a1a2...an=Tn=1-an(1)a1a2...a(n-1)=Tn-1=1-a(n-1)(2)(1)/(2)an=(1-an)/[1-a(n-1)]整理