Working on an assignment I came across the following problem:
Show explicitly the Bianchi identity: $$ R^{a}_{b[cd;e]} = 0 $$ where ; denotes covariant differentiation $$ R^{a}_{bcd} = \frac{1}{2}g^{ae}(\partial_b \partial_cg_{ed}-\partial_d\partial_bg_{ec}+\partial_e\partial_d g_{bc}-\partial_e\partial_cg_{bd}) $$ However I am having trouble expanding the tensor since I do not know how to apply the notation to the LHS.
There are two steps to be carried out. First the anti-symmetrization $[cd;e]$ means you consider every permutation of the indices, and multiply by the sign of the permutation. In this case we have,
$$R^a_{bcd;e} -R^a_{bce;d} + R^a_{bec;d}-R^a_{ebc;d} + R^a_{ebd;c}-R^a_{bed;c} = 0$$
where I have ignored the $\frac{1}{n!}$ factor ($n=\# \, \mathrm{indices}$) since we are setting it all to zero. Now each term with the $;$ notation means, e.g.,
$$R^a_{bcd;e}=\nabla_e R^a_{bcd}.$$
We can expand this further, in terms of partial derivaties and Christoffel symbols:
$$\nabla_e R^a_{bcd} = \partial_e R^a_{bcd} + \Gamma^a_{ef}R^f_{bcd} - \Gamma^f_{eb}R^a_{fcd} - \Gamma^f_{ec}R^a_{bfd}-\Gamma^f_{ed} R^a_{bcf}.$$
You can notice the pattern, we contract each index with $\Gamma$, putting the missing index on $\Gamma$, a $+$ for contravariant indices and $-$ for a covariant index.
Can you take it from here? To prove the Bianchi identity it suffices to plug in the definition of the Riemann tensor, though this is not the least tedious method. It's a quick few lines if you use differential forms.
