$x(n)={1,2,3,4}$

If $x(n) (↔)^{DFT} X(k)$ & $y(n)(↔)^{DFT} Y(k)$then

$∑_{n=0}^{N-1}x(n).y^* (n)=\frac{1}{N} ∑_{n=0}^{N-1}X(k).Y^* (k)$

Proof:-We have

$r_{xy} (m)=∑_{n=0}^{N-1}x(n).y^* (n-m)_N$

At

$m=0 ,r_{xy} (m)=∑_{n=0}^{N-1}x(n).y^* (n)$ ………(1)

By DFT, we have

$DFT{r_{xy} (m) }=X(K).Y^* (k)$

$r_{xy} (m)=IDFT{X(k).Y^* (k)}$

By IDFT eqn

$r_{xy} (m)=\frac{1}{N} ∑_{k=0}^{N-1}X(k).Y^* (K).e^{\frac{j2πkm}{N}}$

at m=0

$r_{xy} (0)=\frac{1}{N} ∑_{k=0}^{N-1}X(k).Y^* (k)$ ……………(2)

By comparing eq (1)& (2)

$∑_{n=0}^{N-1}x(n).y^* (n)=\frac{1}{N} ∑_{k=0}^{N-1}X(k).Y^* (k)$

$∑_{n=0}^{N-1}(x(n))^2=\frac{1}{N} ∑_{k=0}^{N-1}(|X(k) |)^2$

Given x(n)={1,2,3,4}

Consider LHS

$∑_{n=0}^{N-1}(|x(n) |)^2=∑_{n=0}^3(|x(n) |)^2$

$=(1)^2+(2)^2+(3)^2+(4)^2$

=30

Consider RHS

$\frac{1}{N} ∑_{k=0}^{N-1}(|X(k) |)^2 =\frac{1}{4} ∑_{k=0}^3(|X(k) |)^2$

Calculate DFT of x(n)

$X(K)={10,-2+2j,-2,-2-2j}$

$=\frac{1}{4} [(10)^2+(-2+2j)^2+(-2)^2+(-2-2j)^2 ]$

$=\frac{1}{4} {100+(+8)+(4)+(8)}$

$=\frac{1}{4}[100+8+4+8]$

$=30$