百度作业网一道 设数列{an}是等差数列.{bn}是各项为正等比数列a1=b1=1,a3+b5=21,a5+b3=131) 求an
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一道
设数列{an}是等差数列.{bn}是各项为正等比数列
a1=b1=1,a3+b5=21,a5+b3=13
1) 求an、bn
2)求数列{an/bn}的前n项和
设数列{an}是等差数列.{bn}是各项为正等比数列
a1=b1=1,a3+b5=21,a5+b3=13
1) 求an、bn
2)求数列{an/bn}的前n项和
1)a1=b1=1,a3+b5=21,a5+b3=13
则1+2d+q^4=21,1+4d+q^2=13,
得2q^4-q^2-28=0.
可知q^2=4,d=2.
所以,an=2n-1,bn=2^(n-1)(因为bn为正).
2)S=1+3/2+5/4+7/8+9/16+……+(2n-1)/[2^(n-1)],
(1/2)S=1/2+3/4+5/8+7/16+9/32+……+(2n-3)/[2^(n-1)]+(2n-1)/(2^n),
(1/2)S=1+2{1/2+1/4+1/8+1/16+……+1/[2^(n-1)]}-(2n-1)/(2^n)
=1+4[1-(1/2)^(n-1)]-(2n-1)/(2^n)
=5-(1/2)^(n-3)-(2n-1)/(2^n),
S=10-(1/2)^(n-4)-(4n-2)/(2^n).
则1+2d+q^4=21,1+4d+q^2=13,
得2q^4-q^2-28=0.
可知q^2=4,d=2.
所以,an=2n-1,bn=2^(n-1)(因为bn为正).
2)S=1+3/2+5/4+7/8+9/16+……+(2n-1)/[2^(n-1)],
(1/2)S=1/2+3/4+5/8+7/16+9/32+……+(2n-3)/[2^(n-1)]+(2n-1)/(2^n),
(1/2)S=1+2{1/2+1/4+1/8+1/16+……+1/[2^(n-1)]}-(2n-1)/(2^n)
=1+4[1-(1/2)^(n-1)]-(2n-1)/(2^n)
=5-(1/2)^(n-3)-(2n-1)/(2^n),
S=10-(1/2)^(n-4)-(4n-2)/(2^n).
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