Maclaurin series stuck at finding $L_n$

Question

I need to develop Maclaurin serie of $f(x)=\frac{1}{(1-x)^2}$

I found all the derivative, and all the zero values for the derivatives. I come up with that :

$\frac{1}{(1-x)^2}=1+2x+\frac{6}{2!}x^2+\frac{24}{3!}x^3+\frac{120}{4!}x^4+...+\frac{?}{(n-1)!}x^{n-1} + ...$

Now I want to determinate $L_n=\frac{f^{n-1}(0)}{(n-1)!}x^{n-1}$

And this is where I'm stuck. How can I calculate $L_n$ ?

Answer 1

When calculating successive derivatives, sometimes it is more illuminating to not simplify out the entire calculation: $$\begin{align*} f(x) &= (1-x)^{-2}, \\ f'(x) &= 2(1-x)^{-3}, \\ f''(x) &= 2(3)(1-x)^{-4}, \\ f'''(x) &= 2(3)(4)(1-x)^{-5}, \\ f^{(4)}(x) &= 2(3)(4)(5)(1-x)^{-6}, \\ &\vdots \\ f^{(n)}(x) &= \ldots. \end{align*}$$ Does the pattern seem more evident now?

Answer 2

Start from the geometric series $$\frac{1}{1-x}=\sum\limits_{n=0}^{\infty}{x^{n}} $$ add differentiate it:$$ \frac{d}{dx}{\left(\frac{1}{1-x} \right)}=\frac{1}{(1-x)^2}=\sum\limits_{n=1}^{\infty}{nx^{n-1}}.$$

Answer 3

$$\frac{ 1 }{ 1-x }=1+x+x^2+x^3+x^4+x^5+...\\\frac{ 1 }{ 1-x }\frac{ 1 }{ 1-x }=\\(1+x+x^2+x^3+x^4+x^5+...)(1+x+x^2+x^3+x^4+x^5+...)\\\\=1+(x+x)+(x^2+x^2+xx)+(x^3+x^3+x^2x+xx^2)+...=\\\\1+2x+3x^2+4x^3+5x^4+6x^5+... $$