2.an=2/(n+1)
a1=1及a(n+1)=2an/(2+an),得:
a1=1
a2=2/3
a3=1/2=2/4
a4=2/5
猜测:an=2/(n+1)
证明:
1、当n=1时,an=a1=2/(1+1)=1,满足;
2、设:当n=k时,ak=2/(k+1)
则当n=k+1时,
a(k+1)=2ak/(2+ak)
=2/[(k+1)+1]
即当n=k+1时也成立
3.V+F-E=2
4.n=1时,2^n-1= 1,(n+1)^2 = 4,2^n-1=8 时,都有 2^n-1 >(n+1)^2