还有一道数列的题已知数列{an}和{bn}都是等差数列,数列{an}的前n项和记作Sn,数列{bn}的前n项和记作Tn,且Sn/Tn=(2n-3)/(4n-3),求(1)(a2000)/(b2000)(2)[(b100)/(a1999+a1)]+[(b1900)/(a1900+a100)]
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还有一道数列的题已知数列{an}和{bn}都是等差数列,数列{an}的前n项和记作Sn,数列{bn}的前n项和记作Tn,且Sn/Tn=(2n-3)/(4n-3),求(1)(a2000)/(b2000)(2)[(b100)/(a1999+a1)]+[(b1900)/(a1900+a100)]
还有一道数列的题
已知数列{an}和{bn}都是等差数列,数列{an}的前n项和记作Sn,数列{bn}的前n项和记作Tn,且Sn/Tn=(2n-3)/(4n-3),求
(1)(a2000)/(b2000)
(2)[(b100)/(a1999+a1)]+[(b1900)/(a1900+a100)]
还有一道数列的题已知数列{an}和{bn}都是等差数列,数列{an}的前n项和记作Sn,数列{bn}的前n项和记作Tn,且Sn/Tn=(2n-3)/(4n-3),求(1)(a2000)/(b2000)(2)[(b100)/(a1999+a1)]+[(b1900)/(a1900+a100)]
Sn=n(a1+an)/2 Tn=n(b1+bn)/2
Sn/Tn=(a1+an)/(b1+bn)=2n-3/4n-3
⑴:a2000=1/2(a1+a3999)
b2000=1/2(b1+b3999)
所以,a2000/b2000=(a1+a3999)/(b1+b3999)=(2×3999-3)/(4×3999-3)=8995/15993
⑵:a1999+a1=2a1000
a1900+a100=2a1000
所以,原式=(b100+b1900)/2a1000
=2b1000/2a1000=(b1+b1999)/(a1+a1999)=(4×1999-3)/(2×1999-3)=7993/1995