作业帮已知anbn都是等差数列,其中n项和分别为sntn,若sn/tn=n/2n+1
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已知anbn都是等差数列,其中n项和分别为sntn,若sn/tn=n/2n+1
求a5/b5.
求sn/bn
求a5/b5.
求sn/bn
a5/b5=[(a1+a9)/2]/[(b1+b9)/2] /分子分母同时运用等差中项性质
=[(a1+a9)×9/2]/[(b1+b9)×9/2]
=S9/T9
=9/(2×9+1)
=9/19
第二问是an/bn吧.
an/bn=[(a1+a(2n-1))/2]/[(b1+b(2n-1))/2]
=[(a1+a(2n-1))(2n-1)/2]/[(b1+b(2n-1))(2n-1)/2]
=S(2n-1)/T(2n-1)
=(2n-1)/[2(2n-1)+1]
=(2n-1)/(4n-1)
如果是Sn/bn,那么：
Sn/Tn=[na1+n(n-1)d1/2]/[nb1+n(n-1)d2/2]
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=[d1n+(2a1-d1)]/[d2n+(2b1-d2)]
=n/(2n+1)
令d1=t (t≠0),则2a1-d1=0 d2=2t 2b1-d2=t
解得a1=t/2 b1=3t/2 d1=t d2=2t
Sn/bn=[na1+n(n-1)d1/2]/[b1+(n-1)d2]
=[n(t/2)+n(n-1)(t/2)]/[(3t/2)+(n-1)(2t)]
=[n+n(n-1)]/[3 +4(n-1)]
=n²/(4n-1)
=[(a1+a9)×9/2]/[(b1+b9)×9/2]
=S9/T9
=9/(2×9+1)
=9/19
第二问是an/bn吧.
an/bn=[(a1+a(2n-1))/2]/[(b1+b(2n-1))/2]
=[(a1+a(2n-1))(2n-1)/2]/[(b1+b(2n-1))(2n-1)/2]
=S(2n-1)/T(2n-1)
=(2n-1)/[2(2n-1)+1]
=(2n-1)/(4n-1)
如果是Sn/bn,那么：
Sn/Tn=[na1+n(n-1)d1/2]/[nb1+n(n-1)d2/2]
=[2a1+(n-1)d1]/[2b1+(n-1)d2]
=[d1n+(2a1-d1)]/[d2n+(2b1-d2)]
=n/(2n+1)
令d1=t (t≠0),则2a1-d1=0 d2=2t 2b1-d2=t
解得a1=t/2 b1=3t/2 d1=t d2=2t
Sn/bn=[na1+n(n-1)d1/2]/[b1+(n-1)d2]
=[n(t/2)+n(n-1)(t/2)]/[(3t/2)+(n-1)(2t)]
=[n+n(n-1)]/[3 +4(n-1)]
=n²/(4n-1)
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