Question

等差数列前几项和SnTn,Sn/Tn=2n/3n+1求an/bn

Answer 1

等差数列前几项和SnTn,Sn/Tn=2n/3n+1
则an/bn=[a1+a(2n-1)]/2÷[b1+b(2n-1)]/2={[a1+a(2n-1)]/2}×(2n-1)÷{[b1+b(2n-1)]/2}×(2n-1)
=S(2n-1)÷T(2n-1)=2(2n-1)÷[3(2n-1)+1]=(4n-2)/(6n-2)=(2n-1)/(3n-1)

Answer 2

设an=a1+(n-1)d1/2,bn=b1+(n-1)d2/2
Sn=(a1+an)n/2
Tn=(b1+bn)n/2
Sn/Tn=2n/(3n+1)=[a1+an]/[b1+bn]
S(2n-1)/T(2n-1)=2(2n-1)/(6n-2)=[a1+a(2n-1)]/[b1+b(2n-1)]
=[a1+a1+(2n-2)d1]/[b1+b1+(2n-2)d2]
=[a1+(n-1)d1]/[b1+(n-1)d2]
=an/bn
an/bn=(2n-1)/(3n-1)

Answer 3

a1/b1=2/3+1=5/3 a1=5x b1=3x Sn=n*(a1+an)/2 Tn=n*(b1+bn)/2 Sn/Tn=5n/3n
Sn=5ny Tn=3ny Sn=5ny=na1/2+nan/2 an=10y-5x bn=6y-3x an/bn=5*(2y-x)/3*(2y-x)=5/3
an/bn=2/3+1