已知Sn,Tn,且Sn Tn=7n-3 n 1,求a5 b5

S15=(a1+a15)*15/2T15=(b1+b15)*15/2所以S15/T15=(a1+a15)/(b1+b15)等差数列,则a8和b8是a1,a15以及b1,b15的等差中项所以a1+a15

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/

因为但看1+2+3...+n这个数列,通项公式为n(n+1)/2=n^/2+n/2所以1=1/2(1^+1)1+2=1/2(2^+2)1+2+3=1/2(3^+3)以此类推,提出共因数1/2,合并括号

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）

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

∵anbn＝2an2bn＝a1+a2n−1b1+b2n−1=(2n−1)(a1+a2n−1) 2(2n−1)(b1+b2n−1) 2＝s2n−1T2n−1∴anbn=2(2n−1)

由等差数列的性质Sn=na1+n(n-1)d/2=dn2/2+(a1-d/2)n=An2+Bn即A=d/2B=a1-d/2同样地Tn=nb1+n(n-1)p/2=pn2/2+(b1-p/2)n=Cn2

{an}是等差数列,a2=a1+da3=a1+2d....an=a1+(n-1)da(2n-1)=a1+(2n-2)da1+a(2n-1)=2a1+(2n-2)d2an=2a1+2(n-1)d=2a1

4/3S21=(a1+a21)*21/2a1+a21=a1+a1+20d=2(a1+10d)=2a11所以S21=2a11*21/2=21*a11同理T21=21*b11所以a11/b11=S21/T

设等差数列{an}和{bn}的公差分别为d1 和d2，则由题意可得S1T1＝a1b1＝2×13×1+1=12，即2a1=b1．再由S2T2＝a1+a2b1+b2＝2a1+d12b1+d2＝2

因为等差数列前n项和为Sn=na1+n(n-1)d/2=d/2*n^2+(a1-d/2)*n所以可知等差数列前n项和是关于n的二次函数,且不含常数项.因为Sn/Tn=(7n+45)/(n+3),所以可

等我算算啊,几分钟

S(2n-1)=(2n-1)an,T(2n-1)=(2n-1)an,所以an/bn=S(2n-1)/T(2n-1),所以a9/b9=S17/T17=18/31.

∵等差数列{an}、{bn}，∴an=a1+a2n−12，bn=b1+b2n−12，∴anbn=nannbn=n(a1+a2n−1)2n(b1+b2n−1)2=S2n−1T2n−1，又SnTn=7n+

T（n+1）-Tn=a（n+1）=1-a（n+1）-1+an,即a（n+1）=an/2.T1=1-a1,得a1=1/2.∴an是首项为1/2公比为1/2的等比数列,得an=（1/2）ⁿ,同

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

∵等差数列{an}和{bn}的前n项和分别为Sn和Tn，且SnTn＝5n+32n+7，a5b5=9a59b5=s9T9=4825故选B．

S21=(a1+a21)*21/2a1+a21=a1+a1+20d=2(a1+10d)=2a11所以S21=2a11*21/2=21*a11同理T21=21*b11所以a11/b11=S21/T21=

S9=(a1+a9)/2*9=(2a5)/2*9=9a5同理T9=9b5a5／b5=S9:T9=21/4

∵{an},{bn}是两个等差数列∴a1+a15=2a8b1+b15=2b8∴a8/b8=(15(a1+a15)/2)/(15(b1+b15)/2)=S15/T15∵Sn/Tn=(7n+2)/(n+3