Transformations solving recurrence with generating functions

Question

Why is

• $\sum_{n ≥3} a_{n-1} z^{n-1}$ equal to $(A(z) − 1 − z)$?
• $\sum_{n ≥3}a_{n-2} z^{n-2}$ equal to $(A(z) − 1)$?

Answer 1

$$\sum_{n\geqslant3}a_{n-1}z^{n-1} = \sum_{n\geqslant2}a_nz^n = A(z) - a_0 - a_1z = A(z) - 1 -z. $$

$$\sum_{n\geqslant3}a_{n-2}z^{n-2} = \sum_{n\geqslant1}a_nz^n = A(z) - a_0 = A(z) - 1. $$