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*}
Best Answer
You certainly do not need generating functions for solving $a_n-a_{n-1}=2(n-1)$: it is enough to sum both sides on $n=1,2,\ldots,N$ to get that $a_n$ depends on $\sum_{k=1}^{n}2(k-1)$. The only issue in your case is that the given values of $a_0$ and $a_1$ do not agree with the given recurrence relation. So, let us assume $a_0=\color{red}{2}$ and $a_{n}=a_{n-1}+2(n-1)$ for any $n\geq 1$. By denoting as $f(x)$ the following generating function $$ f(x)=\sum_{n\geq 0} a_n x^n = 2+\sum_{n\geq 1}a_n x^n$$ we have $$ x\cdot f(x) = \sum_{n\geq 0} a_n x^{n+1} = \sum_{n\geq 1} a_{n-1} x^n $$ hence by the recurrence relation $$ (1-x) f(x) = 2+2\sum_{n\geq 1}(n-1)x^n = 2+\frac{2x^2}{(1-x)^2}$$ where the last identity follows from stars and bars in the form $$ \frac{1}{(1-x)^{m+1}}=\sum_{n\geq 0}\binom{n+m}{m}x^n.$$ Since we have $$ f(x) = \frac{2}{1-x}+\frac{2x^2}{(1-x)^3} $$ stars and bars also gives $$ a_n = [x^n]f(x) = 2-n+n^2. $$