假设Xn+k=a[X(n-1)+k]Xn+k=aX(n-1)+akXn=aX(n-1)+ak-k与Xn=aX(n-1)+b比较可得ak-k=b,k=b/(a-1)所以Xn+[b/(a-1)]=a[X(n-1)+(b/(a-1))]数列{Xn+b/(a-1)}是公比为a的等比数列,首项是X0+[b/(a-1)]所以其通项Xn+[b/(a-1)]...