# Finding the convolution $a^n u[n]$ with $b^n u[n]$

Hi I'm trying to solve the problem when studying for an upcoming test. The given solution is $$y[n]= \frac{1}{b-a}(b^{n+1}-a^{n+1}) \quad \text{for } n\ge0 \ .$$ However, I'm not sure how to reach this point. We have to also verify using the z-transform. Any help would be appreciated so much.

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