已知等差数列{an}与{bn}的前n项和分别为Sn与Tn,(1)Sn/Tn=(n-13)/(4n+7)

显示全部楼层 · 2012-2-13 19:18:10

解：(1)设an=a1+(n-1)d1,bn=b1+(n-1)d2，则Sn=na1+n(n-1)/2*d1Tn=nb1+n(n-1)/2*d2Sn/Tn=(n-13)/(4n+7)=[na1+n(n-1)/2*d1]/[nb1+n(n-1)/2*d2]=[a1+(n-1)/2*d1]/[b1+(n-1)/2*d2]=(nd1+2a1-d1)/(nd2+2b1-d2)=[n+(2a1/d1-1)]/[d2/d1*n+(2b1-d2)/d1]=(n-13)/(4n+7)于是有2a1/d1-1=-13d2/d1=4(2b1-d2)/d1=7得：a1=-6d1b1=11/2*d1d2=4d1则a6/b10=(...

千问 · 2012-2-13 19:18:10

a1/b1=S1/T1=(-12)/11设a1=-12t, 则b1=11t则a(n) 公差为d,b(n)的公差为h,则a(n+1)=a1+nd=-12t+ndb(n+1)=b1+nh=11t+nhS(n+1)/T(n+1)=[a1+a(n+1)]/[b1+b(n+1)]=[nd-12t]/[nh+11t]=[n-12]/[4n+1...