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Let : $$ x[n]=\begin{cases} 1&\text{if $0\leq n\leq 4$}\\ \\ 0&\text{if otherwise} \end{cases} \qquad \text{and} \qquad h[n]=\begin{cases} \alpha^{n}&\text{if $0\leq n\leq 6$}\\ \\ 0&\text{if otherwise} \end{cases} $$ We must compute the convolution : $$ (x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k] $$ Observe that $x[n]=u[n]$ for $n\in[0,4]$ and $0$ otherwise. Furthermore, $h[n]=\alpha^{n}u[n]$ for $n\in[0,6]$. Thus : \begin{array}{c | c c c c c c c } \mathbf{x} & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ \hline 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 0 & 0 & 0 & 0& 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0& 0& 0 & 0 & 0 \end{array} \begin{align*} \displaystyle (x*h)[n]&=\left[\sum_{k=0}^{0}\alpha^{k}\;,\;\sum_{k=0}^{1}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=0}^{4}\alpha^{k}\;,\;\sum_{k=1}^{5}\alpha^{k}\;,\;\sum_{k=2}^{6}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=5}^{6}\alpha^{k}\;,\;\sum_{k=6}^{6}\alpha^{k}\right]\\ \\ &=\begin{cases} \displaystyle\sum_{k=0}^{n}\alpha^{k}&\text{if $0\leq n\leq 4$}\\ \displaystyle\sum_{k=1}^{n}\alpha^{k}&\text{if $n=5$}\\ \displaystyle\sum_{k=n+2}^{6}\alpha^{k}&\text{if $0\leq n\leq4$} \end{cases} \end{align*} I wish to know where my mistake is and how to fix it. Thank you :)

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$3^{rd}$ term in your final answer should have $6 \le n \le 10$, instead of $0 \le n \le 4$. And, the limit of summation will go from $n-4$ to $6$.

$$\sum^{6}_{k = n-4} \alpha^k \ , 6 \le n \le 10$$

$0 \le n \le 4$ case has already been covered by the $1^{st}$ term of final answer.

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