已知a1=2,an不等于0,且an+1-an=2an+1an,求an

a[n+1]-a[n]=2a[n+1]a[n]
1/a[n]-1/a[n+1]=2
1/a[n+1]=(1/a[n])-2
1/a[n]为等差数列,公差为-2,首项1/a[1]=1/2
所以 1/a[n]=1/2-2(n-1)=5/2-2n=(5-4n)/2
a[n]=2/(5-4n)