等差数列前几项和SnTn,Sn/Tn=2n/3n+1求an/bn
人气:284 ℃ 时间:2020-03-24 17:58:47
解答
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)
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