<s>[INST] Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not?
 [/INST] There are many excellent proofs given above. 
Another approach would be to note that the summation (LHS) is actually a binomial expansion of the result (RHS). (NB - This would only be applicable if you have to prove that the result is true, instead of having to find the result.)
Hence
$$\begin{align}
\color{blue}{\frac 1{1-x}}&=(1-x)^{-1}\\
&=\sum_{r=0}^{\infty}{-1\choose r}(-x)^r\\
&=\color{lightgray}{1+(-1)(-x)+\frac{-1\cdot-2}{1\cdot 2}(-x)^2+\frac{-1\cdot -2\cdot -3}{1\cdot 2\cdot 3}(-x)^3\cdots}\\
&=\sum_{r=0}^{\infty}(-1)^r{r\choose r}(-x)^r
\qquad\text{using upper negation}\\
&=\color{green}{\sum_{r=0}^{\infty}{r\choose 0}x^r} =\sum_{r=0}^{\infty}x^r \\
&=1+x+x^2+x^3+\cdots+x^r+\cdots \\
&\phantom{=}\text{(Geometric Series)}
\end{align}$$
and 
$$\begin{align}
\color{blue}{\frac 1{(1-x)^2}}&=(1-x)^{-2}\\
&=\sum_{r=0}^{\infty}{-2\choose r}(-x)^r\\
&=\color{lightgray}{1+(-2)(-x)+\frac{-2\cdot-3}{1\cdot 2}(-x)^2+
\frac{-2\cdot-3\cdot -4}{1\cdot 2\cdot 3}(-x)^3\cdots}\\
&=\sum_{r=0}^{\infty}(-1)^r{r+1\choose r}(-x)^r
\qquad\text{using upper negation}\\
&=\color{green}{\sum_{r=0}^{\infty}{r+1\choose 1}x^r}=\sum_{r=0}^{\infty}(r+1)x^r \\
&=1+2x+3x^2+4x^3+\cdots+(r+1)x^r+\cdots \\
&\phantom{=}\text{(Arithmetico-Geometric Series)}\\
\end{align}$$

Similarly, 
$$\begin{align}
\color{blue}{\frac 1{(1-x)^3}}&=(1-x)^{-3}\\
&=\sum_{r=0}^{\infty}{-3\choose r}(-x)^r\\
&=\color{lightgray}{1+(-3)(-x)+\frac{-3\cdot-4}{1\cdot 2}(-x)^2+
\frac{-3\cdot-4\cdot -5}{1\cdot 2\cdot 3}(-x)^3\cdots}\\
&=\sum_{r=0}^{\infty}(-1)^r{r+2\choose r}(-x)^r
\qquad\text{using upper negation}\\
&=\color{green}{\sum_{r=0}^{\infty}{r+2\choose 2}x^r}=\sum_{r=0}^{\infty}\frac{(r+2)(r+1)}2x^r \\
&=1+3x+6x^2+10x^3+\cdots+\frac{(r+2)(r+1)}2 x^r+\cdots \\
\end{align}$$
This can also be derived by differentiating the result of the Arithmetico-Geometric Series.
By extension, 
$$\color{blue}{\frac 1{(1-x)^{n+1}}}=\color{green}{\sum_{r=0}^{\infty}{r+n\choose n}x^r}$$
</s>