Suppose I have the power series $\frac{3}{2} \sum_{n=0}^{\infty} (\frac{x}{2})^{n}$ with center $c=0$. Suppose I wanted to change it to have the center $c=-3$. Is the following legal?
$\frac{3}{2} \sum_{n=0}^{\infty} (\frac{x}{2})^{n}$ becomes $\frac{3}{2} \sum_{n=0}^{\infty} (\frac{x+3}{2})^{n}$ (could multiply by 2 to find the new center $c=3$.
Is this allowed/the right way?
No, what you are asked to do is to find coefficients $a_n$ such that $\frac 3 2 \sum\limits_{k=0}^{\infty} (\frac x 2)^{k}=\sum\limits_{k=0}^{\infty} a_k (x+3)^{k}$. For this note that $\frac 3 2 \sum\limits_{k=0}^{\infty} (\frac x 2)^{k}=\frac 3 2 \frac 1 {1-\frac x 2}=\frac 3 {2-x}$. Now $\frac 3 {2-x}=\frac 3 {5-(x+3)}=\frac 3 5 \sum\limits_{k=0}^{\infty} [\frac {(x+3)} 5]^{k}$. This is the required expansion around $x=-3$.
[$a_k=\frac 3 {5^{k+1}}$].
