1. **Problem statement:** Find the output $y[n]$ of a system with impulse response $h[n] = \left(\frac{1}{6}\right)^n u[n]$ when the input is $x[n] = (n+1)\left(\frac{1}{4}\right)^n u[n]$ using the Fourier transform. 2. **Recall the system output formula:** The output $y[n]$ is the convolution of $x[n]$ and $h[n]$: $$y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k]$$ 3. **Fourier transform approach:** The convolution in time domain corresponds to multiplication in frequency domain: $$Y(e^{j\omega}) = X(e^{j\omega}) H(e^{j\omega})$$ 4. **Find $H(e^{j\omega})$:** For $h[n] = a^n u[n]$ with $a=\frac{1}{6}$, the DTFT is $$H(e^{j\omega}) = \sum_{n=0}^\infty \left(\frac{1}{6}\right)^n e^{-j\omega n} = \frac{1}{1 - \frac{1}{6} e^{-j\omega}}$$ 5. **Find $X(e^{j\omega})$:** For $x[n] = (n+1) b^n u[n]$ with $b=\frac{1}{4}$, use the known formula: $$\sum_{n=0}^\infty (n+1) r^n = \frac{1}{(1-r)^2}$$ Replacing $r = b e^{-j\omega}$, $$X(e^{j\omega}) = \sum_{n=0}^\infty (n+1) \left(\frac{1}{4}\right)^n e^{-j\omega n} = \frac{1}{\left(1 - \frac{1}{4} e^{-j\omega}\right)^2}$$ 6. **Calculate $Y(e^{j\omega})$:** $$Y(e^{j\omega}) = X(e^{j\omega}) H(e^{j\omega}) = \frac{1}{\left(1 - \frac{1}{4} e^{-j\omega}\right)^2} \cdot \frac{1}{1 - \frac{1}{6} e^{-j\omega}}$$ 7. **Inverse DTFT to find $y[n]$:** Recognize that $$\frac{1}{(1 - r e^{-j\omega})^m} \leftrightarrow \frac{(n+m-1)!}{n! (m-1)!} r^n u[n]$$ 8. **Express $Y(e^{j\omega})$ as product of two terms:** $$Y(e^{j\omega}) = \frac{1}{(1 - \frac{1}{4} e^{-j\omega})^2} \cdot \frac{1}{1 - \frac{1}{6} e^{-j\omega}}$$ 9. **Convolution in time domain:** This corresponds to convolution of sequences $$y[n] = x_1[n] * x_2[n]$$ where $$x_1[n] = (n+1) \left(\frac{1}{4}\right)^n u[n], \quad x_2[n] = \left(\frac{1}{6}\right)^n u[n]$$ 10. **Compute convolution:** $$y[n] = \sum_{k=0}^n (k+1) \left(\frac{1}{4}\right)^k \left(\frac{1}{6}\right)^{n-k}$$ 11. **Simplify:** $$y[n] = \left(\frac{1}{6}\right)^n \sum_{k=0}^n (k+1) \left(\frac{3}{2}\right)^k$$ 12. **Sum evaluation:** The sum $$S_n = \sum_{k=0}^n (k+1) r^k$$ with $r=\frac{3}{2}$ has closed form $$S_n = \frac{1 - (n+2) r^{n+1} + (n+1) r^{n+2}}{(1-r)^2}$$ 13. **Final expression:** $$y[n] = \left(\frac{1}{6}\right)^n \cdot \frac{1 - (n+2) \left(\frac{3}{2}\right)^{n+1} + (n+1) \left(\frac{3}{2}\right)^{n+2}}{\left(1 - \frac{3}{2}\right)^2} u[n]$$ 14. **Simplify denominator:** $$1 - \frac{3}{2} = -\frac{1}{2}, \quad \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$$ 15. **Therefore:** $$y[n] = 4 \left(\frac{1}{6}\right)^n \left[1 - (n+2) \left(\frac{3}{2}\right)^{n+1} + (n+1) \left(\frac{3}{2}\right)^{n+2}\right] u[n]$$ This is the output of the system for the given input.