Closed loop equation initial value
You're trying to form an iterative equation. There are many ways of going about it, sometimes it works, sometimes it doesn't.

This way works:

$$\small V_o=\frac{A}{1+AB}V_i $$

add \$\small V_o\$ to both sides:

$$\small 2V_o=\frac{A}{1+AB}V_i +V_o$$

Divide by 2:
$$\small V_o=\frac{1}{2}\left(\frac{A}{1+AB}V_i +V_o\right)$$
Make it iterative:
$$\small V_o[k]=\frac{1}{2}\left(\frac{A}{1+AB}V_i +V_o[k-1]\right)$$

Let \$\small A=1000; B=0.25; V_i=1\$
$$\small V_o[k]=\frac{1}{2}\left(\frac{1000}{251} +V_o[k-1]\right)$$
hence:

$$\small V_o[k]=\frac{500}{251} +\frac{V_o[k-1]}{2}$$

Let \$\small V_o[0]=0\$, then solve iteratively:

$$\small V_o[1]=1.992$$
$$\small V_o[2]=2.988$$
$$\small V_o[3]=3.486$$
$$\small V_o[4]=3.735$$
This converges to \$\small V_o[k]=3.984\$ at 17 iterations.

Interestingly(?), a square root algorithm can be formulated in a similar way:

Let \$\small y=\sqrt{x}\$ 
$$\small y^2=x$$
$$\small y=\frac{x}{y}$$
$$\small 2y=\frac{x}{y}+y$$
$$\small y=\frac{1}{2}\left(\frac{x}{y}+y\right)$$
Make iterative:
$$\small y[k]=\frac{1}{2}\left(\frac{x}{y[k-1]}+y[k-1]\right)$$

e.g. let \$\small x=9,\:y[0]=1\$
$$\small y[1]=\frac{1}{2}\left(\frac{9}{1}+1\right)=5$$
$$\small y[2]=\frac{1}{2}\left(\frac{9}{5}+5\right)=3.4$$
$$\small y[3]=\frac{1}{2}\left(\frac{9}{3.4}+3.4\right)=3.0235$$
$$\small y[4]=\frac{1}{2}\left(\frac{9}{3.0235}+3.0235\right)=3.00009$$