Solution

The equation are
$$2x+ky-k=0$$
$$4x+2y-(k+1)=0$$
Here, $$ a_{1}=2,b_{1}=k,c_{1}=-k$$
and $$ a _{2}=4,b_{2}=2,c_{2}=-(k+1)$$
For the system to have infinite solutions,
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{2}{4}=\frac{k}{2}=\frac{-k}{-(k+1)}$$

Taking,
$$\displaystyle \frac{2}{4}=\frac{k}{2}$$

$$\Rightarrow 4k=4$$

$$\Rightarrow k=1$$

Taking,
$$\displaystyle\frac{k}{2}=\frac{-k}{-(k+1)}$$

$$\Rightarrow k+1=2 \Rightarrow k=1$$

So, $$k=1$$ is the answer