an为等差数列 bn为等比数列a1=b1=2,a2-b2=1,a3=b3问(1)求an和bn的通项公式(2)设cn=an/bn(n为正整数),数列cn的前n项和为Tn,求证Tn小于5
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![an为等差数列 bn为等比数列a1=b1=2,a2-b2=1,a3=b3问(1)求an和bn的通项公式(2)设cn=an/bn(n为正整数),数列cn的前n项和为Tn,求证Tn小于5](/uploads/image/z/3620520-0-0.jpg?t=an%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+bn%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97a1%3Db1%3D2%2Ca2-b2%3D1%2Ca3%3Db3%E9%97%AE%EF%BC%881%EF%BC%89%E6%B1%82an%E5%92%8Cbn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%BEcn%3Dan%2Fbn%EF%BC%88n%E4%B8%BA%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%2C%E6%95%B0%E5%88%97cn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82%E8%AF%81Tn%E5%B0%8F%E4%BA%8E5)
an为等差数列 bn为等比数列a1=b1=2,a2-b2=1,a3=b3问(1)求an和bn的通项公式(2)设cn=an/bn(n为正整数),数列cn的前n项和为Tn,求证Tn小于5
an为等差数列 bn为等比数列a1=b1=2,a2-b2=1,a3=b3
问(1)求an和bn的通项公式
(2)设cn=an/bn(n为正整数),数列cn的前n项和为Tn,求证Tn小于5
an为等差数列 bn为等比数列a1=b1=2,a2-b2=1,a3=b3问(1)求an和bn的通项公式(2)设cn=an/bn(n为正整数),数列cn的前n项和为Tn,求证Tn小于5
1.
A2=2+d A3=2+2d
B2=2q B3=2q^2
2+d-2q=1 1+d=2q
2+2d=2q^2 1+d=q^2
解方程得q=2,d=3 q=0舍去,等比数列公比不为0
An=3n-1
Bn=q^n
2.
Tn=2/2^1+5/2^2+8/2^3+……+(3n-1)/2^n
2Tn=2/2^0+5/2^1+8/2^2+……+(3n-1)/2^(n-1)
2Tn-Tn=2+3[1/2^1+1/2^2+1/2^3+……1/2^(n-1)]-(3n-1)/2^n
=2+3/2×(1-(1/2)^(n-1)/(1-(1/2))-(3n-1)/2^n
=5-(3n+5)/2^n
n>=1时
3n+5>0
2^n>0
(3n+5)/2^n>0
所以Tn
解:(1)设an=2+d(n-1);bn=2q^(n-1)
所以 2+2d-d-2q=1 解得 d=3 q=2 所以an=3n-1 bn=2^n
2+3d-d=2q^2
(2)cn=3n-1/2^n
Tn=(3*1-1)*2^(-1)+(3*2-1)*2^(-2)+(3*3-1)*2^(-3)+...+(3n-1)*2^(-n)①
Tn/2=...
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解:(1)设an=2+d(n-1);bn=2q^(n-1)
所以 2+2d-d-2q=1 解得 d=3 q=2 所以an=3n-1 bn=2^n
2+3d-d=2q^2
(2)cn=3n-1/2^n
Tn=(3*1-1)*2^(-1)+(3*2-1)*2^(-2)+(3*3-1)*2^(-3)+...+(3n-1)*2^(-n)①
Tn/2= (3*1-1)*2^(-2)+(3*2-1)*2^(-3)+...+(3n-4)*2^(-n)+(3n-1)*2^(-n-1)②
②-①得 Tn/2=1-(3n-1)*2^(-n-1)+3*(2^(-2)+2^(-3)+...+2^(-n)
由此可得 Tn=5-(3n+5)/2^n<5
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