Prove that $\nabla_\mu G^{\mu\nu} = 0$ I have Einstein's Equation

$$\large G_{\mu\nu} = 8\pi GT_{\mu\nu}$$

And I know that when I have a vector say $S^\sigma$ then

$$\large \nabla_\mu S^\sigma = \partial_\mu S^\sigma + \Gamma^\sigma_{\mu\lambda}S^\lambda$$

But I am not sure what
$$\large \nabla_\mu G^{\mu\nu}$$
would be with two indices. Would anyone mind showing me?

EDIT:
Would it look something like this?
$$\nabla_\mu G^{\mu\nu} = \partial_\mu G^{\mu\nu} + \Gamma^{\mu}_{\alpha\mu}G^{\alpha\nu} + \Gamma^{\nu}_{\alpha\mu}G^{\mu\alpha}$$
And if so then why would these three terms become zero?
 A: What you are trying to do is called covariant derivative. Indeed you need to write a term with Christoffel symbol for each index. For covariant derivative of a tensor field, you do:
$$
\nabla_\mu T_{kl}^{ij} = \partial_\mu T_{kl}^{ij} + \Gamma^{i}_{\alpha\mu}T_{kl}^{\alpha j} + \Gamma^{j}_{\alpha\mu}T_{kl}^{i\alpha}
-\Gamma^{\alpha}_{k\mu}T_{\alpha l}^{ij}
-\Gamma^{\alpha}_{l\mu}T_{k\alpha}^{ij}.
$$
Note sign $+$ for contravariant (upper) indices and sign $-$ for covariant (lower) indices.
After you have written the contravariant derivative, you can contract it on the desired indices.
Your expression in EDIT has several index problems (you should have only one upper index $\nu$ in each term, all other indices should go in pairs, one top, one bottom).
$$
\nabla_\mu G^{\mu\nu} = \partial_\mu G^{\mu\nu} + \Gamma^{\mu}_{\alpha\mu}G^{\alpha\nu} + \Gamma^{\nu}_{\alpha\mu}G^{\mu\alpha}$$
A: The answer of Vasily is true and is enough. This is just an explanation:$$\Large{\nabla_\mu T_{\color{purple}k\color{green}l}^{\color{red}i\color{blue}j} = \partial_\mu T_{kl}^{ij} + \color{red}{\Gamma^{i}_{\alpha\mu}T_{kl}^{\alpha j}} + \color{blue}{\Gamma^{j}_{\alpha\mu}T_{kl}^{i\alpha}}
-{\color{purple}{\Gamma^{\alpha}_{k\mu}T_{\alpha l}^{ij}}}
-\color{green}{\Gamma^{\alpha}_{l\mu}T_{k\alpha}^{ij}}.}$$
