Can the expression $\sum_{r=1}^{n}\sum_{i=0}^{n-r} {{n-r} \choose {i}}a^{r+i-1}b^{n-(r+i)}$ be further simplified?

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I am trying to teach myself about summations and techniques of simplifying a complicated expression. I was wondering if the following summation can be simplified, or maybe possibly re-written in a different form?

$$\sum_{r=1}^{n}\sum_{i=0}^{n-r} {{n-r} \choose {i}}a^{r+i-1}b^{n-(r+i)}$$

I have tried expanding the term ${{n-r}\choose i}$, but it has not gotten me to anything meaningful.

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$$\begin{align}\sum_{r=1}^{n}\sum_{i=0}^{n-r} {{n-r} \choose {i}}a^{r+i-1}b^{n-(r+i)}&= \sum_{r=1}^{n}a^{r-1}\sum_{i=0}^{n-r} {{n-r} \choose {i}}a^{i}b^{(n-r)-i)} \\ &= \sum_{r=1}^{n}a^{r-1}(a+b)^{n-r}\\ &= (a+b)^{n-1}\sum_{r=1}^{n}\left(\frac{a}{a+b}\right)^{r-1}\\ &=(a+b)^{n-1}\dfrac{1-\left(\frac a{a+b}\right)^n}{1-\frac{a}{a+b}}\\ &=\frac{(a+b)^n-a^n}{b} \end{align}$$