1、a(n+1)=2an+1
a(n+1)+1=2an+2=2[an+1]
则:[a(n+1)+1]/[an+1]=2=常数,则数列{an+1}是以an+1=2为首项、以q=2为公比的等比数列,则:an+1=2×2^(n-1)=2^n,则:an=2^n-1;
2、a(n+1)=2an/[2+an] 两边取倒数,得:
1/[a(n+1)]=[2+an]/(2an)=1/an+(1/2) 则:
1/[a(n+1)]-1/[an]=1/2=常数,则数列{1/an}是以1/a1=1为首项、以d=1/2为公差的等差数列,则:1/[an]=1+(1/2)(n-1)=(n+1)/2,则:an=2/(n+1)