I don't really understand how to solve (with generating functions) for the recurrence relation of $$a_n = a_{n-1}+2(n-1)$$ with initial conditions of $a_1 = 2$ when $n \geq 2$
This is what I was thinking, \begin{align*} g(x) &= a_0x^0+a_1x^1+a_2x^2+...+a_rx^r+...\\ g(x) &= a_0 + a_1x^1 + \sum^{\infty}_{n=2}a_nx^n\\ g(x) &= 1 + 2x + \sum^{\infty}_{n=2}{(a_{n-1}+2(n-1))x^n}\\ g(x) &= 1 + 2x + \sum^{\infty}_{n=2}a_{n-1}x^n + \sum^{\infty}_{n=2}(2n-2)x^n\\ g(x) &= 1 + 2x + x\sum^{\infty}_{n=1}a_{n-1}x^{n-1} + \sum^{\infty}_{n=2}2n x^n - \sum^{\infty}_{n=2}2x^n\\ g(x) &= 1 + 2x + x\sum^{\infty}_{m=1}a_{m}x^{m} + \sum^{\infty}_{n=2}2 \binom{n}{1}x^n - \sum^{\infty}_{n=2}2x^n\\ g(x) &= 1 + 2x + x(g(x) - a_0) + \sum^{\infty}_{n=2}2 \binom{n}{1}x^n - \sum^{\infty}_{n=2}2x^n\\ \end{align*}