Proof of Lyndon-Hochschild-Serre spectral

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I am checking about the Lyndon-Hochschild-Serre spectral sequence, which given the next group extension

$1\rightarrow H\rightarrow G\rightarrow Q\rightarrow 1$ we have the next spectral sequence

$E_{p,q}^{2}=H_{p}(Q,H_{q}(H,M))\Rightarrow H_{p+q}(G,M)$

for $M$ a $G-$module. I am trying to find the proof of the construction of this spectral sequence so that I can see the terms $E_{p,q}^{0}$ and $E_{p,q}^{1}$, so Where can I find that construction? any guide will be a bleasing-

Thank you for your time and consideration!

----------------------Edit--------------------------

I have been searching and I found this, in the book of Cohomology of groups by Kenneth Brown given a double complex in general $C=(C_{p,q})_{p,q\in \mathbb{Z} }$ with horizontal differential $\partial^{h}$ and vertical differential $\partial^{v}$ both of degree $-1$; i.e.

$\partial^{h}:C_{p,q}\rightarrow C_{p-1,q}$

and

$\partial^{v}:C_{p,q}\rightarrow C_{p,q-1}$

Now we can define a spectral sequence (two depending of the filtration that is used) where

$E_{p,q}^{0}=C_{p,q}$ with differential $d^{0}=\pm \partial^{v}$

$E_{p,q}^{1}=H_{q}(C_{p,\ast})$ with $d^{1}$ with differential induced by $\partial^{v}$

On group homology we recall that $H_{\ast}(G,M)=H_{\ast}(F\otimes_{G} M)$ where $F$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}[G]$ and $M$ a $\mathbb{Z}[G]-$module. This can be generalized by a nonnegative complex $C=(C_{n})_{n\geq0}$ by

$H_{\ast}(G,C)=H_{\ast}(F\otimes_{G} C)$.

Thus for the double complex $F\otimes_{G} C$ we have a spectral sequence where

$E_{p,q}^{0}=F_{p}\otimes_{G} C_{q}$

$E_{p,q}^{1}=H_{q}(F_{p}\otimes_{G} C_{\ast})=F_{p}\otimes_{G} H_{q}(C_{\ast})$ (The last equality follows fron the exactness of the functor $F_{p}\otimes_{G}-$)

From this for the split group extension $1\rightarrow H\rightarrow G\rightarrow Q\rightarrow 1$ and $F$ a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}[G]$ we have that

$F\otimes_{G}M=((F\otimes M)_{H})_{Q}=(F\otimes_{H}M)_{Q}$

Thus

$H_{\ast}(G,M)=H_{\ast}(C_{Q})$

where $C=F\otimes_{H}M$, now taking $\widetilde{F}$ a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}[Q]$ we can work with the double complex $\widetilde{F}\otimes_{Q}(F\otimes_{H}M)$ so that we can have the Lyndon-Hochschild-Serre spectral sequence but in this case we have

$E_{p,q}^{0}=\widetilde{F}_{p}\otimes_{Q}(F_{q}\otimes_{H}M)$

and

$E_{p,q}^{1}=\widetilde{F}_{p}\otimes_{G} H_{q}(H,M)$

Now I can see the elements of the zeroth page and first page. I have a question, in the construction we use a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}[G]$.

Can I use a a resolution over $\mathbb{Z}[H]$?