已知数列｛an｝的首项a1=9,且a(n+1)=2an/1+an,求该数列通项an,

Question

已知数列｛an｝的首项a1=9,且a(n+1)=2an/1+an,求该数列通项an,

Answer 1

a(n+1)=2a(n)/[1+a(n)],
若a(n+1)=1,则2a(n)=1+a(n),a(n)=1,...,a(1)=1与a(1)=9矛盾.
因此,a(n)不为1.
a(n+1)-1 = [2a(n)-1-a(n)]/[1+a(n)] = 2[a(n)-1]/[1+a(n)],
1/[a(n+1)-1] = (1/2)[a(n)+1]/[a(n)-1] = (1/2)[a(n)-1+2]/[a(n)-1]=1/[a(n)-1] + 1/2
{1/[a(n)-1]}是首项为1/[a(1)-1]=1/8,公差为1/2的等差数列.
1/[a(n)-1] = 1/8 + (n-1)/2 = [4(n-1)+1]/8 = (4n-3)/8,
a(n)-1=8/(4n-3),
a(n)=1+8/(4n-3)=(4n+5)/(4n-3)