记等比数列an的前n项积为Tn,若a3=-1.T5等于

等比数列{an}的公比为q,其前n项的积为Tn,∵a1＞1,a99a100-1＞0,a99=a1*q^98,a100=a1*q^99∴（a1)^2*q^(197)>1那么q>0,an>0,若q>1,又

（1）证：Sn=S1+a2[1-(-1/2)^(n-1)]/(1-(-1/2))=S1-(1/3)a1[1-(-1/2)^(n-1)]≤S1,当n=1时,等号成立Sn=S2+a3[1-(-1/2)^(

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

(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

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+

首项为a1,公比为q则an=a1*q^(n-1)所以1/a1+1/a2+1/a3+.+1/an=1/a1*[(1+1/q+1/q^2+...+1/q^(n-1)]=1/a1*[(1-1/q^n)/(1

(a2006-1)(a2007-1)1,soqa20070T(n)=中间项的n次方=>T(n)与1的关系,中间项和1的关系,是一致的a2007是第一项小于1的a2007是T(4013的中间项)----

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

因为S1=a1=(2^1)-1=1；又：Sn=a1*(1-q^n)/(1-q),且依题得知q=2所以,得an=a1*q^(n-1)=2^(n-1)则bn={(an)^2}=2^(2n-2)在数列bn中

数列{an}、{bn}分别为正项等比数列,数列{lgan}{lgbn}是等差数列Tn/Rn=n/2n+1则假设Tn=k*n^2,Rn=k*n*(2n+1)k>0lgan=k*(2n-1)lga5=9k

（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−

an=1536(-1/2)^(n-1)T(n+1)=-1/2an*TnT(n+1)1536[(-1/2)^n]*TnTn一定为正,化简（-1/2）^n*1536

1.Tn=1-An=1-Tn/T(n-1)两边除以Tn:1=1/Tn-1/T(n-1)1/Tn-1/T(n-1)=1Cn-C(n-1)=1则Cn是首项为1/(1-A1),公差为1的等差数列.2.Tn=

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*

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^