Why $R_n=n^2$ for sum-recurrence and solving for $\alpha=0, \beta=-1, \gamma=2$

Question

Starting from these definitions from the book Concrete Mathematics,

The author continues to the highlighted text below

I don't follow

1. How the choice for $R_n=n^2$ was made
2. How the author reached $\alpha=0, \beta=-1, \gamma=2$
3. Is $C(n)=(n^2+n)/2$ a solution for a recurrence of the form (2.8) only because $R_n=n^2$? Would we have found this for $C(n)$ had we not chosen $R_n=n^2$? Furthermore, the solutions for $R_n=n, R_n=n^2$ were found for different particular values of $\alpha, \beta, \gamma$. Why do these solutions for $B(n), C(n)$ hold for all $\alpha, \beta, \gamma$?

Clarification would be helpful.

Answer 1

1. The choice $R_n=n^2$ is because $n^2$ is the next "simple function" of $n$ after $1$ and $n$.
2. From (2.7), $\alpha=R_0=0$ and $\beta+\gamma n=R_n-R_{n-1}=n^2-(n-1)^2=-1+2n$, so $\beta=-1$ and $\gamma=2$.
3. Any quadratic polynomial of $n$ would lead to the same $C(n)$. For an arbitrary example, try $R_n=2n^2-1$. You might also want to read this related post: Mathematical explanation for the Repertoire Method