1. **State the problem:** Given the discrete-time signal $x[n] = 2n u[n-3]$, where $u[n-3]$ is the unit step function shifted by 3, find its discrete-time Fourier transform (DTFT) $X(e^{j\omega})$. 2. **Recall the DTFT formula:** The DTFT of a discrete-time signal $x[n]$ is defined as $$ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}. $$ Since $x[n] = 0$ for $n < 3$ due to the unit step $u[n-3]$, the summation limits reduce to $n=3$ to $\infty$. 3. **Write the summation explicitly:** $$ X(e^{j\omega}) = \sum_{n=3}^{\infty} 2n e^{-j\omega n}. $$ 4. **Simplify the summation:** Factor out the constant 2: $$ X(e^{j\omega}) = 2 \sum_{n=3}^{\infty} n e^{-j\omega n}. $$ 5. **Use the formula for the sum of $n r^n$ from $n=N$ to $\infty$:** For $|r|<1$, $$ \sum_{n=N}^{\infty} n r^n = r^N \frac{N - (N-1)r}{(1-r)^2}. $$ Here, $r = e^{-j\omega}$ and $N=3$. 6. **Apply the formula:** $$ \sum_{n=3}^{\infty} n e^{-j\omega n} = e^{-j3\omega} \frac{3 - 2 e^{-j\omega}}{(1 - e^{-j\omega})^2}. $$ 7. **Substitute back:** $$ X(e^{j\omega}) = 2 e^{-j3\omega} \frac{3 - 2 e^{-j\omega}}{(1 - e^{-j\omega})^2}. $$ **Final answer:** $$ \boxed{X(e^{j\omega}) = \frac{2 e^{-j3\omega} (3 - 2 e^{-j\omega})}{(1 - e^{-j\omega})^2}}. $$