Sum of factorials in 3D symmetrical random walk $\sum_{k=0}^n\binom{n}{k}^2\binom{2(n-k)}{n-k}$

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I'm trying to prove that in a tridimensional symmetrical random walk all states are transient. Since it's an irreducible Markov chain, I only have to prove it for a generic state, since they are all equal. I'm trying to prove it by proving that the series

$$\sum_{n=0}^{+\infty} P^{(n)}<+\infty$$

where $P^{(n)}$ is the probability of returning in the same state after $n$ steps. This probability is zero if $n$ is odd, and for even values I found

$$P^{(2n)}=\sum_{n_x+n_y+n_z=n}\dfrac{(2n)!}{(n_x!)^2(n_y!)^2(n_z!)^2}\left( \dfrac{1}{6}\right)^{2n}$$

After a few calculations I arrived at this formula

$$P^{(2n)}=\left( \dfrac{1}{6}\right)^{2n}\binom{2n}{n}\sum_{k=0}^n\binom{n}{k}^2\binom{2(n-k)}{n-k}$$

Does anybody know how to calculate this sum, or a better way of calculating the original sum?