设等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,若Tn\Sn=4n+27\7n+1,求bn\an
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设等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,若Tn\Sn=4n+27\7n+1,求bn\an
![设等差数列{an}的前n项和为Sn,等差数列{bn}的前n项和为Tn,若Tn\Sn=4n+27\7n+1,求bn\an](/uploads/image/z/15884308-28-8.jpg?t=%E8%AE%BE%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%8B%A5Tn%5CSn%3D4n%2B27%5C7n%2B1%2C%E6%B1%82bn%5Can)
设等差数列{an}的公差为a
等差数列{bn}的公差为b
S(2n-1)=[a1+a(2n-1)]*(2n-1)/2
=2an*(2n-1)/2
=(2n-1)an
T(2n-1)=[b1+b(2n-1]*(2n-1)/2
=2bn*(2n-1)/2
=(2n-1)bn
S(2n-1)/T(2n-1)=[(2n-1)an]/[(2n-1)bn]
=an/bn
∴an/bn=[4(2n-1)+27]/{7(2n-1)+1]=(8n+23)/(14n-6)
等差数列{bn}的公差为b
S(2n-1)=[a1+a(2n-1)]*(2n-1)/2
=2an*(2n-1)/2
=(2n-1)an
T(2n-1)=[b1+b(2n-1]*(2n-1)/2
=2bn*(2n-1)/2
=(2n-1)bn
S(2n-1)/T(2n-1)=[(2n-1)an]/[(2n-1)bn]
=an/bn
∴an/bn=[4(2n-1)+27]/{7(2n-1)+1]=(8n+23)/(14n-6)
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