两个等差数列{an}{bn}前n项和SnTn,若Sn Tn=2n 3n 1

an,（bn)^2,a(n+1)成等差数列2(bn)^2=an+a(n+1)--①由（bn）^2,a(n+1),(b(n+1))^2成等比数列(a(n+1))^2=[bnb(n+1)]^2∴a(n+1

S9/T9=9a5/9b5=a5/b5=63/12=21/4S8/T8=4(a4+a5)/[4(b4+b5)]=(a4+a5)/(b4+b5)=56/11S7/T7=7a4/7b4=a4/b4=49/

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

1.S2n+1=(A1+A2n+1)*(2n+1)/2=(2n+1)*An（由等差中项推导出来）,同理T2n+1=(2n+1)*Bn.所以An/Bn=S2n+1/T2n+1=（4n+4）/（2n+3）

S(2n-1)=(A1+A(2n-1))×(2n-1)/2=(A1+A1+(2n-2)d)×(2n-1)/2=(A1+(n-1)d)×(2n-1)=An×(2n-1)同理T(2n-1)=Bn×(2n-

由AnBn＝7n+45n+3，可设An=kn（7n+45）⇒an=An-An-1=14kn+38k，设Bn=kn（n-3）⇒bn=Bn-Bn-1=2kn+2k，所以a2n=28kn+38k，a2nbn

Sn=n(A1+An)/2Tn=n(B1+Bn)/2Sn/Tn=(A1+An)/(B1+Bn)然后n代2n-1A2n-1+A1=2AnBn同理S2n-1/T2n-1=An/Bn=7(2n-1)/(2n

题目都说是猜了所以先找规律a1=1b1=2an,bn,an+1成等比数列a2=4bn,an+1,bn+1成等差数列b2=6依次得到a3=9b3=12a4=16b4=20...可以看出an=n^2bn=

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

A11/B11=（33-3）/（22+3）=6/5A11=11/2(a1+a11)=11/2(2a1+10d)=11/2(2a6)=11a6A11=a1*11+11(11-1)d/2=11/2(2a1

令n=9，得到S9T9＝7×9+29+3=6512，又S9=9(a1+a9) 2=9a5，T9=9(b1+b9) 2=9b5，∴S9T9=9a59b5=a5b5=6512．故答案为

由等差数列的通项公式可得a2+a5+a17+a22b8+b10+b12+b16=2(2a1+21d)2(2b1+21d′)=a1+a22b1+b22=22(a1+a22)222(b1+b22)2=S2

设Sn=k(7n^2+n)an=Sn-S(n-1)=k(14n-6)Tn=k(4n^2+27n)bn=Tn-T(n-1)=k(8n+23)an:bn==(14n-6)/(8n+23)再问：错·再答：哪

∵SnTn＝7n+3n+3∴a8b8＝2a82b8=a1+a15b1+b15=152(a1+a15)152(b1+b15)=S15T15=7×15+315+3=6故答案为：6

∵SnTn=n2n+1，∴a7b7=2a72b7=132(a1+a13)132(b1+b13)=S13T13=132×13+1=1327，故选：C．

由题意可得S14T14=14(a1+a14)214(b1+b14)2=2a72b7=a7b7=3×14+24×14−5=4451，故答案为：4451．

解析,Sn和Tn是an和bn的前n项和,因此,Sn/Tn=(7n+3)/(n+3)=[n(7n+3)]/[n(n+3)]=(7n²+3n)/(n²+3n)设Sn=k(7n²

{an}和{bn}公差分别设为d1、d2Sn=na1+n(n-1)d1/2Tn=nb1+n(n-1)d2/2Sn/Tn=[2a1+(n-1)d1]/[2b1+(n-1)d2]=(2n+3)/(3n-1

A9＝9a5(A9=a1+a2+...a9)(因为a1+a9=2a5,a2+a8=2a5,a3+a7=2a5,a4+a6=2a5)同样的道理,B9＝9b5所以a5/b5＝A9/B9＝28/21=4/3

等差数列数列的性质a1+a[2n-1]=2an因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)anT[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1