I am not really sure how to compute this with Maple, but here is how to get it in closed form.
Recall that the negative binomial distribution with parameters $n$ and $p$ has the following simple probability generating function: $ \mathcal{P}_X\left(z\right) = \left( \frac{p}{1-(1-p)z} \right)^n $ And that the probability generating function for the sum of two independent random variables $X_1+X_2$ is a product of individual generating functions: $ \mathcal{P}_{X_1+X_2}(z) = \left( \frac{p_1}{1-(1-p_1)z} \right)^{n_1} \left( \frac{p_2}{1-(1-p_2)z} \right)^{n_2} $ The probability mass function for $X_1+X_2$ is then read off as a series coefficient of PGF: $ \mathbb{P}\left(X_1+X_2 = n\right) = [z^n] \mathcal{P}_{X_1+X_2}(z) $ In particular: $ \mathbb{P}\left(X_1+X_2=0\right) = p_1^{n_1} p_2^{n_2} = \mathbb{P}\left(X_1=0\right) \mathbb{P}\left(X_2=0\right) $
By the way, here is what Mathematica gives for your problem:
In[14]:= Simplify[ PDF[TransformedDistribution[ x1 + x2, {Distributed[x1, NegativeBinomialDistribution[2, 1/2]], Distributed[x2, NegativeBinomialDistribution[6, 3/10]]}], k]] Out[14]= Piecewise[{{243*2^(-14 - k)*5^(-7 - k)*(66*5^(7 + k) - 15030*7^(3 + k) + 3*5^(7 + k)*k + 30673*7^(2 + k)*k - 4300*7^(2 + k)*k^2 + 60*7^(3 + k)*k^3 - 20*7^(2 + k)*k^4 + 2*7^(2 + k)*k^5), k >= 0}}, 0]
Compare with the series coefficient:
In[18]:= With[{p1 = 1/2, p2 = 3/10}, Simplify[SeriesCoefficient[(p1^2* p2^6)/((1 + (-1 + p1)*z)^2*(1 + (-1 + p2)*z)^6), {z, 0, k}]]] Out[18]= Piecewise[{{243*2^(-14 - k)*5^(-7 - k)*(66*5^(7 + k) - 15030*7^(3 + k) + 3*5^(7 + k)*k + 30673*7^(2 + k)*k - 4300*7^(2 + k)*k^2 + 60*7^(3 + k)*k^3 - 20*7^(2 + k)*k^4 + 2*7^(2 + k)*k^5), k >= 0}}, 0]