I have been trying to attempt this question for quite a while and I am not sure how to approach it. Would anyone be able to help out with the working?
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Hint
The simplest way is to use the Z transform property "convolution in time domain is multiplication in z domain". See Z transform convolution
$$\mathrm{Z}(x[n]*h[n]) = \mathrm{Z}(x[n]) \times \mathrm{Z}(h[n])$$
Then you just need to do inverse Z transform.
Some typical Z transforms (including what is useful for you) can be found at Z transform pairs
Be careful with region of convergence (ROC).
It seems a homework?
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First replace $x[k]$ and $h[n-k]$ with their respective expressions in $y[n]$, then take out of the sum anything that does not depend on $k$, then try simplifying what you have inside your sum. You should get something of the form:
$$y[n]=\alpha\sum_{k=-\infty}^{+\infty}\beta[k]u[k]u[n-k-3].$$
The limits of the sum (lets call them $l_1$ and $l_2$) are determined by the product $u[k]u[n-k-3]$ (where the product is nonzero). You should then obtain something of the form:
$$y[n]=\alpha\sum_{k=l_1}^{l_2}\beta[k].$$
Hint: $\beta[k]$ should be of the form $\gamma^k$, where $\gamma <1$ and the sum is of a geometric series.
