Deriving the perturbative expansion from Hubbard to Heisenberg

[hokhani]

TL;DR Summary

I can not go one step further in this expansion.

In the youtube lecture “electron interaction and the Hubbard model” at the time 2:23:00, we have the following self-consistent equation with energy appearing at both sides:
$$(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P) |\phi \rangle_{g.s.} =E |\phi \rangle_{g.s.}$$
Where $H_0$ is the unperturbed Hamiltonian, $H_1$ the perturbation, $|\phi \rangle_{g.s.}$ is the ground state ket of the full Hamiltonian $(H_0+H_1)$, and $\hat P (\hat Q)$ is the projection operator on the ground (excited) states of $H_0$.
By defining the effective Hamiltonian as:
$$H_{eff}=(\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P)$$
The self-consistent equation is as:
$$H_{eff}|\phi \rangle_{g.s.}=E|\phi \rangle_{g.s.}$$
So, my question:
How does the solution of this effective Hamiltonian, recursively, give the following equation?
$$H_{eff}=\hat P \hat H_0 \hat P+\hat P \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P+\hat P \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat Q (E_0-\hat Q \hat H_0 \hat Q)^{-1} \hat Q \hat H_1 \hat P +$$
Where the first, second and third lines are respectively zero, second and third order terms.

I would be grateful if anyone could please provide any help with that.