fx=x/x+2,数列an满足a1=1,an+1=f(an)求数列an的通项公式若数列bn满足bn=2^nanan+1,Sn=b1+b2+…+bn,求证Sn

Question

fx=x/x+2,数列an满足a1=1,an+1=f(an)求数列an的通项公式
若数列bn满足bn=2^n an an+1,Sn=b1+b2+…+bn,求证Sn

Answer 1

a(n+1)=f(an)=an/(an+2)
取倒数得到1/a(n+1)=1+2/an
即有1/a(n+1)+1=2[1/an+1]
故数列{1/an+1}是一个首项是1/a1+1=2,公比是2的等比数列,则有:
1/an+1=2*2^(n-1)=2^n
故有an=1/(2^n-1)
2.
bn = 2^n.an.a(n+1)
= 2^n/[(2^n-1)(2^(n+1) -1)]
= 1/(2^n-1) -1/(2^(n+1) -1)
Sn=b1+b2+...+bn = 1/(2-1) - 1/(2^(n+1) -1)
= 1- 1/(2^(n+1) -1)
Sn < 1