<s>[INST] Expansion of this expression Let $x$ be a real number in $\left[0,\frac{1}{2}\right].$ It is well known that 
$$\frac{1}{1-x}=\sum_{n=0}^{+\infty} x^n.$$
What is the expansion or the series of the expression $(\frac{1}{1-x})^2$?
Many thanks.
 [/INST] Method 1 : Binomial Theorem for any index
$$(1-x)^{-n}= 1+nx+ \frac{n(n+1)}{2!}x^2+ \frac{n(n+1)(n+2)}{3!}x^3 \dots $$
Putting $n=1$ , we get :
$$\frac{1}{1-x} = 1+x+x^2+x^3+ \dots $$
Putting $n=2$ , we get :
$$\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$
Method 2 : Derivatives
Let: $$f(x)=\frac{1}{1-x} = 1+x+x^2+x^3+ \dots $$
Take derivative w.r.t. $x$ both the sides;
$$f'(x)=\frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$
$$ \implies \frac{1}{(1-x)^2} = 1+2x+3x^2+4x^3+ \dots $$
NOTE : These expansions are valid not only for $ x \in \left[0,\dfrac{1}{2}\right]$ but $\forall ~x \in (-1,1)$
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