设可化为a[n+1]+x=p(a[n]+x)²+y,则x=q/(2p),y=r+x-px²作换元b[n]=p(a[n]+x),则b[n+1]=b[n]²+py若py≠0,则b[n]=((b[1]²+py)²+py)²+.) (共有n-1个括号)上式是不能化简的.若py=0,即y=0,...