Solution:

The inverse DFT is defined as,

$ x(n)=\frac{1}{N} \sum_{k=0}^{N-1} X(k) e^{j 2 \pi n k / N}, n=0,1,2,3, \ldots, N-1\\ $

$ \begin{aligned} & \text { Given } N=4, x(n)=\frac{1}{4} \sum_{k=0}^3 X(k) e^{j 2 \pi n k / N}, n=0,1,2,3 \\\\ & \text { When } n=0 \\\\ & x(0)=\frac{1}{4} \sum_{k=0}^3 X(k) e^{j \pi(0) k / 2} \\\\ & =\frac{1}{4}(1+2+3+4)=\frac{5}{2} \\\\ & \end{aligned} $

When $n=1$

$ \begin{aligned} x(1) & =\frac{1}{4} \sum_{k=0}^3 X(k) e^{j \pi(1) k / 2} \\\\ & =\frac{1}{4}\left(1+2 e^{j \pi / 2}+3 e^{j \pi}+4 e^{j 3 \pi / 2}\right) \\\\ & =\frac{1}{4}(1+2(j)+3(-1)+4(-j)) \\\\ & =\frac{1}{4}(-2-j 2)=-\frac{1}{2}-j \frac{1}{2}\\ \end{aligned}\\ $

When $n=2$

$ \begin{aligned} x(2) & =\frac{1}{4} \sum_{k=0}^3 X(k) e^{j \pi k} \\\\ & =\frac{1}{4}\left(1+2 e^{j \pi}+3 e^{j 2 \pi}+4 e^{j 3 \pi}\right) \\\\ & =\frac{1}{4}(1+2(-1)+3(1)+4(-1)) \\\\ & =\frac{1}{4}(-2)=-1 / 2\\ \end{aligned} $

When $n=3$

$ \begin{aligned} x(3) & =\frac{1}{4} \sum_{k=0}^3 X(k) e^{j 3 \pi k / 2} \\\\ = & \frac{1}{4}\left(1+2 e^{j 3 \pi / 2}+3 e^{j 3 \pi}+4 e^{j 9 \pi / 2}\right)\\ \end{aligned} $

$ \begin{aligned} = & \frac{1}{4}(1+2(-j)+3(-1)+4 j) \\\\ & =\frac{1}{4}(-2+2 j)=-\frac{1}{2}+j \frac{1}{2} \\\\ x(n) & =\left\{\frac{5}{2},-\frac{1}{2}-j \frac{1}{2},-\frac{1}{2},-\frac{1}{2}+j \frac{1}{2}\right\}\\ \end{aligned} $