Concept of finding forced response from difference equation?
considering your equation,

$$k\times u[n] - 0.25\times k\times u[n-2] = 2\times u[n] + u[n-1]$$

the left half of the equation will result in a series as shown below:

$$\begin{array}{cccccccc} [k &amp; k &amp; \frac{3}{4}k &amp; \frac{3}{4}k &amp; \frac{3}{4}k &amp; \frac{3}{4}k  &amp; \ldots &amp;]\end{array}$$

the right hand of the equation is
$$\begin{array}{cccccccc} [2 &amp; 3 &amp; 3 &amp; 3 &amp; 3 &amp; 3 &amp; \ldots &amp; ]\end{array}$$

for these two series to be equal,
$$\begin{array}{cccccccc} k = 2 \\ k=3\\ \frac{3}{4}k=3 \\ \frac{3}{4}k=3 \\ \vdots &amp; \end{array}$$

It is not possible to equate these two sequences without removing first two points. In that case, $k = 4$. The sequences are equal or the equation holds for $k = 4$ and $n \ge 2$. 

What we are getting here is total solution, so it will have the natural response and forced response. The output get stabilized and reach the stable value only after $n \ge 2$. This 'transients' before $n &lt; 2$ are due to the natural response.