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Let $x(n)=\left(\frac{1}{2}\right)^{n} u(n), y(n)=x^{2}(n)$ and $\mathrm{Y}\left(e^{j i e}\right)$ be the Fourier transform of $y(n)$. Then $Y\left(e^{j i e}\right)$ is

1. $\frac{1}{4}$
2. $2$
3. $4$
4. $\frac{4}{3}$