Finding the convolution $a^n u[n]$ with $b^n u[n]$
Just do it the hard way by writing out the individual formulas for the general convolution sum
\begin{align}
y[n] &amp;= \sum_{k=-\infty}^\infty a[k]b[n-k]\\
&amp;= \sum_{k=0}^\infty a[k]b[n-k] &amp;\scriptstyle{\text{because }a[k]=0~\text{whenever } k &lt; 0,}\\
&amp;= \sum_{k=0}^n a[k]b[n-k] &amp;\scriptstyle{\text{because }b[n-k]=0~\text{whenever } k &gt; n,}
\end{align}
which gives
\begin{array}{rclcl}
y[0] &amp;= &amp;a[0]b[0] &amp;= &amp;1\\
&amp;&amp;&amp;= &amp;\displaystyle\frac{b^1-a^1}{b-a}\\
y[1] &amp;= &amp;a[0]b[1] + a[1]b[0] &amp;= &amp;b+a\\
&amp;&amp;&amp;= &amp;\displaystyle\frac{b^2-a^2}{b-a}\\
y[2] &amp;= &amp;a[0]b[2] + a[1]b[1] + a[2]b[0] &amp;= &amp;b^2 + ba +a^2\
&amp;&amp;&amp;= &amp;\displaystyle\frac{b^3-a^3}{b-a}\\
&amp;&amp;&amp;= &amp;\displaystyle\frac{b^3-a^3}{b-a}\\
\vdots &amp;= &amp;\ddots~~~\scriptstyle{\text{Hey, Ma! I think I see a pattern developing here}}\\
\\
y[n] &amp;= &amp;a[0]b[n] + a[1]b[n-1] + \cdots + a[n]b[0]
&amp;= &amp;b^n + b^{n-1}a + \cdots + ba^{n-1} + a^n\\
&amp;&amp;&amp;= &amp;\displaystyle\frac{b^{n+1}-a^{n+1}}{b-a}
\end{array}
if you can recall the formula for the sum of a geometric series.