Recurrence relation chip and conquer
The answer should be $\Theta(3^{n/4})$. You can go both ways you suggest – find an exact formula, or estimate the sum, using an integral or in more devious ways. There is no loss in generality if you take $c = 1$. If you replace $n$ with $1$ (so we get a lower bound), then we have
$$ 1 + 3 + 3^2 + \cdots + 3^{n/4} = \Theta(3^{n/4}), $$
using the formula for summing a geometric series. If you bring back the $n$, the calculation becomes a bit more difficult, but you should still be able to get a matching upper bound:
$$
n + 3(n-4) + 3^2(n-8) + \cdots + 3^{n/4}(0) = O(3^{n/4}).
$$
There are in fact closed formulas for this kind of sum (try Wolfram alpha), but you can also try to bound without them, say by looking at ratios of consecutive terms and comparison to a geometric series:
$$
3^{n/4} \left(0 + 4\frac{1}{3} + 8\frac{1}{9} + 12\frac{1}{27} + \cdots \right) = O(3^{n/4}),
$$
since the infinite series $4\sum_{m=0}^\infty \frac{m}{3^m}$ converges.