So the task says: Do the Maclaurin expansion of the given term: $$f(x) = \frac{x}{(1-x)^2}$$
what I did was: https://ibb.co/19Ddd8d
The way i tried to solve the task was wrong, I know that i should derive and than multiply with x and that would give me the correct Maclaurin expansion.
But why couldn't I square both sides and then multiply with x instead of deriving and than multiplying with x?
Perhaps you should sniff around some texts about generating functions to get more comfortable with these calculations. You can certainly square both sides and multiply by $x$. For example:
$$\frac{1}{1-x} = \sum_{m=0}^{\infty} x^m$$
Squaring both sides yields \begin{align} \frac{1}{(1-x)^2} &= \left(\sum_{m=0}^{\infty} x^m\right)^2\\ &= (1+x+x^2+\cdots )(1+x+x^2+\cdots)\\ &= (1+2x+3x^2+4x^3+\cdots)\\ &= \sum_{m=0}^{\infty} (m+1)x^m. \end{align}
Notice that you'd need to calculate the square of the polynomial. Fortunately it is easy enough in this case.
Finally multiplying by $x$ yields \begin{align} \frac{x}{(1-x)^2} &= \sum_{m=0}^{\infty} (m+1)x^{m+1}\\ &= \sum_{m=1}^{\infty} mx^{m} \end{align}
On the other hand, we can take a derivative and be done with it. Taking the derivative of the right hand side is fairly straightforward, as we just take the derivative of every term individually.
\begin{align} \frac{1}{1-x}\frac{d}{dx} &= \sum_{m=0}^{\infty} x^m\frac{d}{dx}\\ \frac{1}{(1-x)^2} &= \sum_{m=0}^{\infty} mx^{m-1}\\ \frac{x}{(1-x)^2} &= \sum_{m=0}^{\infty} mx^{m}\\ \frac{x}{(1-x)^2} &= \sum_{m=1}^{\infty} mx^{m}\\ \end{align}
I forget if one should really take caution while taking the derivative of a sum $\sum_{m=0}^{\infty} x^m$ as the first term without adjusting indices becomes $0(\frac{1}{x})$, but perhaps someone with more experience can speak to this.