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=
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=
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
接下来该怎么解呢
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
接下来该怎么解呢
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a3=3b3
a1+2d=3a1d²
(3d²-1)a1=2d
a1=2d/(3d² -1)
a5=5b5
a1+4d=5a1d⁴
(5d⁴-1)a1=4d
a1=4d/(5d⁴-1)
2d/(3d²-1)=4d/(5d⁴-1)
5d⁴-6d²+1=0
(d-1)(5d-1)=0
d=1(d≠1,舍去)或d=1/5
a1=2d/(3d² -1)=2/(3d -1/d)=2/(3/5 -5)=-5/11
an=a1+(n-1)d=-5/11 +(1/5)(n-1)=-36/55 +n/5
bn=b1d^(n-1)=(-5/11)×(1/5)^(n-1)=-1/[11×5^(n-2)]
a1+2d=3a1d²
(3d²-1)a1=2d
a1=2d/(3d² -1)
a5=5b5
a1+4d=5a1d⁴
(5d⁴-1)a1=4d
a1=4d/(5d⁴-1)
2d/(3d²-1)=4d/(5d⁴-1)
5d⁴-6d²+1=0
(d-1)(5d-1)=0
d=1(d≠1,舍去)或d=1/5
a1=2d/(3d² -1)=2/(3d -1/d)=2/(3/5 -5)=-5/11
an=a1+(n-1)d=-5/11 +(1/5)(n-1)=-36/55 +n/5
bn=b1d^(n-1)=(-5/11)×(1/5)^(n-1)=-1/[11×5^(n-2)]
1年前
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