因为a(n+1)an + a(n+1) - 2an = 0;
所以a(n+1) = 2an/(an+1);
所以a2=2a1/(a1+1)=4/3;
由题可得:
a1 = 2^1/(2^1-1);
a2=2^2/(2^2-1);
由上可得:a(n+1)=2an/(an+1);
a3=2a2/(a2+1)=8/7=2^3/(2^3-1);
a4=2a3/(a3+1)=16/15=2^4/(2^4-1);
……
以此类推可得an=2^n/(2^n-1).
因为a(n+1)an + a(n+1) - 2an = 0;
所以a(n+1) = 2an/(an+1);
所以a2=2a1/(a1+1)=4/3;
由题可得:
a1 = 2^1/(2^1-1);
a2=2^2/(2^2-1);
由上可得:a(n+1)=2an/(an+1);
a3=2a2/(a2+1)=8/7=2^3/(2^3-1);
a4=2a3/(a3+1)=16/15=2^4/(2^4-1);
……
以此类推可得an=2^n/(2^n-1).