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?

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, 
$$\color{blue}{\frac 1{(1-x)^{n+1}}}=\color{green}{\sum_{r=0}^{\infty}{r+n\choose n}x^r}$$
