1. **Problem:** Determine the z-transform and ROC for the sequence $x[n] = (3n + 2 \times 3^n) u[n-1]$. 2. **Formula:** The z-transform of a sequence $x[n]$ is defined as $$X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}$$ and the ROC is the set of $z$ values for which this sum converges. 3. **Step 1: Express the sequence with unit step:** Since $u[n-1]$ is zero for $n<1$, the sum starts at $n=1$: $$X(z) = \sum_{n=1}^\infty (3n + 2 \times 3^n) z^{-n} = \sum_{n=1}^\infty 3n z^{-n} + 2 \sum_{n=1}^\infty 3^n z^{-n}$$ 4. **Step 2: Compute each sum separately:** - For $\sum_{n=1}^\infty 3n z^{-n}$, write as $3 \sum_{n=1}^\infty n (z^{-1})^n$. - For $\sum_{n=1}^\infty 3^n z^{-n}$, write as $\sum_{n=1}^\infty (3 z^{-1})^n$. 5. **Step 3: Use known series formulas:** - Geometric series: $$\sum_{n=1}^\infty r^n = \frac{r}{1-r}, \quad |r|<1$$ - Weighted sum: $$\sum_{n=1}^\infty n r^n = \frac{r}{(1-r)^2}, \quad |r|<1$$ 6. **Step 4: Apply formulas:** - For $3 \sum n (z^{-1})^n$: $$3 \times \frac{z^{-1}}{(1 - z^{-1})^2} = \frac{3 z^{-1}}{(1 - z^{-1})^2}$$ - For $2 \sum (3 z^{-1})^n$: $$2 \times \frac{3 z^{-1}}{1 - 3 z^{-1}} = \frac{6 z^{-1}}{1 - 3 z^{-1}}$$ 7. **Step 5: Combine:** $$X(z) = \frac{3 z^{-1}}{(1 - z^{-1})^2} + \frac{6 z^{-1}}{1 - 3 z^{-1}}$$ 8. **Step 6: ROC:** - For $\sum n (z^{-1})^n$, ROC is $|z| > 1$. - For $\sum (3 z^{-1})^n$, ROC is $|z| > 3$. - Overall ROC is intersection: $$|z| > 3$$ **Final answer:** $$X(z) = \frac{3 z^{-1}}{(1 - z^{-1})^2} + \frac{6 z^{-1}}{1 - 3 z^{-1}}, \quad \text{ROC}: |z| > 3$$ This completes the solution for the first problem.