Question

已知数列{an}中,a1=a>0,an+1=f(an),其中f(x)=2x/x+1.(1)求a2,a3,a4.(2)猜想

Answer 1

a1=a

a2=2a/(a+1)

a3=2a2/(a2+1)=4a/(3a+1)

a4=2a3/(a3+1)=8a/(7a+1)

a(n+1)=2an/(an+1)

取倒数

1/a(n+1)=(an+1)/2an=1/(2an)+1/2

1/a(n+1)-1=1/(2an)-1/2=1/2(1/an-1)

所以1/an-1等比,q=1/2

1/an-1=(1/a-1)*(1/2)^(n-1)=(1-a)/[a*2^(n-1)]

an=1/[1+(1-a)/[a*2^(n-1)]]=a*2^(n-1)/[a*2^(n-1)-a+1]