已知等差数列｛an｝的公差与等比数列｛bn｝的公比相等,且都等于d（d>0,且d≠1）,又知a1=b1,a3=3b3,a5=5b5,求an,bn
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已知等差数列｛an｝的公差与等比数列｛bn｝的公比相等,且都等于d（d>0,且d≠1）,又知a1=b1,a3=3b3,a5=5b5,求an,bn
已知等差数列｛an｝的公差与等比数列｛bn｝的公比相等,且都等于d（d>0,且d≠1）,又知a1=b1,a3=3b3,a5=5b5,求an,bn
已知等差数列｛an｝的公差与等比数列｛bn｝的公比相等,且都等于d（d>0,且d≠1）,又知a1=b1,a3=3b3,a5=5b5,求an,bn
a3=a1+2d a5=a1+4d
b3=b1*d^2 b5=b1*d^4
a1+2d=b1*d^2 a1+4d=5b1*d^4
2d=3a1*d^2-a1 (1) 4d=a1*d^4-a1 (2)
(2)/(1)
2=(5d^4-1)/(3d^2-1)
5d^4-6d^2+1=0
d^2=1 (舍去) d^2=1/5
d=√5/5 a1=-√5
an=-√5+(n-1)*√5/5
bn=-√5*(√5/5)^(n-1)
a3=a1+2d
=3 b3=3b1*d^2
a5=a1+4d
= 5b5=5b1*d^4
so 5d^4 - 6d^2 + 1 = 0
(5d^2-1)(d^2-1)=0
so d=1/sqrt5
so an=(n-1)/sqrt 5 - sqrt 5
bn = 5/sqrt 5
a3=a1+2d b3=b1*d^2 b5=b1*d^4
a1+2d=b1*d^2 a1+4d=5b1*d^4
2d=3a1*d^2-a1
a1+2d=3a1*d^2
a1+4d=5a1*d^4
两式相减得
2d=5a1*d^4-3a1*d^2
2=5a1*d^3-3a1*d
a1=2/(5d^3-3d)
an=2/(5d^3-3d)+(n-1)d
bn=[2/(5d^3-3d)]*d^(n-1)
http://gzsx.cooco.net.cn/testdetail/197315/