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The following convolution $y(n)=a^nu(n) \ast b^nu(n)$ resulted in:

$y(0) = 1$
$y(1) = a+b$
$y(2) = a^2+ab+b^2$
$y(3) = a^3+a^2b+ab^2+b^3$
$y(4) = a^4+a^3b+a^2b^2+ab^3+b^4$
$\vdots$ $\qquad $ $\qquad $ $\qquad $ $\qquad $ $\vdots$
$y(n) = a^n + a^{n-1}b+\cdots+ab^{n-1}+b^n$

Can anyone tell me how to get a closed form for this sequence? If it exists, does this sequence bear a name?

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    Are you sure about your sequence? $y(0),y(1),y(2))$ (a decreasing and b increasing powers) are following a pattern, but $y(3),y(4)$ and the last seem to be different from the pattern. – 2017-02-28
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    @MrYouMath Thank you. I have fixed it. – 2017-02-28

2 Answers 2

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Hint: $(a^n-b^n)=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-1}+b^{n-1})$

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Try multiplying it by $a-b$, and each term should simplify nicely, to give a closed form for $y(n)$ when $a \neq b$. When $a=b$ on the other hand, you can sum the series directly.

Another way of achieving the same result is to look at it as a geometric sequence with first term $a^n$ and ratio $\frac ba$.