一道数列题目求解an,bn都是等差数列.前n项和分别为Sn和Tn Sn/Tn=(n+2)/(2n+1)a3/b5=?谢谢
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一道数列题目求解
an,bn都是等差数列.
前n项和分别为Sn和Tn
Sn/Tn=(n+2)/(2n+1)
a3/b5=?
谢谢
an,bn都是等差数列.
前n项和分别为Sn和Tn
Sn/Tn=(n+2)/(2n+1)
a3/b5=?
谢谢
设数列{an}公差为d,数列{bn}公差为d'
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]
=[2a1+(n-1)d]/[2b1+(n-1)d']
=[dn+(2a1-d)]/[d'n+(2b1-d')]=(n+2)/(2n+1)
令d=t,则2a1-d=2t,d'=2t,2b1-d'=1,
解得a1=3t/2 d=t b1=3t/2 d'=2t
a3/b5=(a1+2d)/(b1+4d')=(3t/2 +2t)/(3t/2 +8t)=(7t/2)/((19t/2)=7/19
再问: 那个应该是2b1-d'=t吧 谢谢了
Sn/Tn=[na1+n(n-1)d/2]/[nb1+n(n-1)d'/2]
=[2a1+(n-1)d]/[2b1+(n-1)d']
=[dn+(2a1-d)]/[d'n+(2b1-d')]=(n+2)/(2n+1)
令d=t,则2a1-d=2t,d'=2t,2b1-d'=1,
解得a1=3t/2 d=t b1=3t/2 d'=2t
a3/b5=(a1+2d)/(b1+4d')=(3t/2 +2t)/(3t/2 +8t)=(7t/2)/((19t/2)=7/19
再问: 那个应该是2b1-d'=t吧 谢谢了
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