@Arjun sir has mentioned the correct link. We can use the Taylor series also.
We can write the Taylor series about origin for the given generating function as :
$f(x) = a_{0} x^{0} + a_{1} x^{1} +a_{2} x^{2}+a_3 x^3+........= \sum_{n}^{} \frac{f^{(n)}(0)}{n!}x^n$ for $n=0,1,2,3,....$
So, for any ordinary generating function,
sequence is : $ a_n= \frac{f^{(n)}(0)}{n!},$ for $n=0,1,2,3,.........$
Here, $f^{(n)}$ means $n^{th}$ derivative of $f$.
Now here, generating function $f(x)=\frac{1}{(1-x)^2}$ and
$ g(i)= \frac{f^{(i)}(0)}{i!},$ for $i=0,1,2,3,.........$
Now,
$f(x)= \frac{1}{(1-x)^2}$
$f'(x)= \frac{1\times2}{(1-x)^3}$
$f''(x)= \frac{1\times 2\times 3}{(1-x)^4}$
$f'''(x)= \frac{1\times 2\times 3 \times 4}{(1-x)^5}$
$.......$
$f^{(i)}(x)= \frac{1\times 2\times 3 \times .......\times (i+1)}{(1-x)^{i+2}}$ $\Rightarrow f^{(i)}(0)= (i+1)!$
So, $g(i) = \frac{f^{(i)}(0)}{i!} = \frac{(i+1)!}{i!} = i+1$