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Refer to the following picture: enter image description here

How can one jump from ($3.2.30$) to $(3.2.31)?$

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Here the passage:

$\nabla_a R_c^a + \nabla_b R_c^b - \nabla_c R = 0$ (3.2.30)
The a and b are dummy indices, the first two terms are the same.
$2 \nabla_a R_c^a - \nabla_c R = 0$
The covariant derivative $\nabla_c$ can be written as $g_{ac} \nabla^a$
In the first term the index a can be raised in the covariant derivative and lowered in the Ricci tensor, via the metric tensor.
$2 \nabla^a R_{ca} - g_{ac} \nabla^a R = 0$
Dividing by 2, moving the metric tensor in the second term after the covariant derivative (metric tensor compatibility) and collecting after the covariant derivative.
$\nabla^a (R_{ca} - \frac{1}{2} g_{ac} R) = 0$
As per definition of Einstein tensor, exchanging indices in the Ricci tensor (index symmetry) and relabelling the free index.
$\nabla^a G_{ab}$ (3.2.31)

Done.

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Just substitute when in doubt:

$$\nabla^a G_{ac} = \nabla^a\left(R_{ac} - \frac{1}{2}Rg_{ac}\right) =\nabla^a R_{ac} - \frac{1}{2}\nabla_c R $$

So $(3.2.30)$ is:

$$2\nabla^a G_{ac}=0$$

Metric compatibility has been assumed here, of course.

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