I don't understand this example presenting a solution to a differential equation using power series

This is an example from a book.

Solve the differential equation $$(4-x^2)y''+y=0.$$

Assume the solution $y(x)=\sum_{n=0}^{\infty}a_nx^n$ exists, where the series has convergence radius $R > 0$. For $|x|<R$ we have that $$y'=\sum_{n=1}^{\infty}na_nx^{n-1},\space y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}$$ $$\Rightarrow x^2y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n}$$

Pluggin this into the LHS of the equation somehow gives that $$(4-x^2)y''+y=\sum_{n=0}^{\infty}[4(n+2)(n+1)a_{n+2}-n(n-1)a_n +a_n]x^n.$$

The thing that confuses me is the middle term $\sum_{n=0}^{\infty}-n(n-1)a_nx^n$ in the RHS of the equation above which is starting from $n=0$. The term obviously comes from $-x^2y''$ on the LHS. From our derivations above we find that $x^2y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n}$, where we start from $n=2$. Now we suddenly start from $n=0$. Why can we do that?

Starting from $$y''=\sum_{n=2}^\infty n(n-1) a_n x^{n-2},$$ you reindex against $n-2$ so it starts at zero, resulting in $$\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n.$$ Then you multiply by $x^2$; this results in $$\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^{n+2}.$$ (This intermediate form wasn't written down in the presentation above.) Finally you reindex back to $n+2$, resulting in $$\sum_{n=2}^\infty n(n-1) a_n x^n.$$

Then finally in writing the series expansion of the actual equation, you can add in the corresponding terms with $n=0$ and $n=1$ from the expansion of $x^2 y''$ since they are just going to be zero anyway due to the $n(n-1)$ factors. (Also, there is no $0/0$ weirdness from doing this, because the exponent on $x$ is nonnegative throughout.) This is convenient since it keeps the indexing scheme the same in all three terms of the equation.