New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$ 
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.

1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
 A: The effect of multiplication by $1/(1-x)$ to the sequence of coefficients is to calculate partial sums: if the original sequence is $c_0,c_1,\ldots$ then the new one is
$$ d_i = c_0 + \cdots + c_i. $$
The starting point is the sequence $1,0,0,\ldots$. Applying this operator twice, we get
$$
1,0,0,0,0,\ldots \\
1,1,1,1,1,\ldots \\
1,2,3,4,5,\ldots
$$
In this matrix, the first row is given, the first column is constant, and otherwise the value of a cell is the sum of the cell above it and the cell to its left.
I'll let you figure out the connection to Pascal's triangle on your own.
A: Problem: For a given integer $N$, how many integers $n_1$ and $n_2$ larger than or equal to zero are there that satisfy the equation:
$$n_1 + n_2 = N$$
We note that it is the coefficient of $x^N$ in the expansion of
$$\left(\sum_{k=0}^{\infty}x^k\right)^2 = \frac{1}{(1-x)^2}$$
We can also solve the problem by noting that it is the number of ways you can color N objects with 2 colors. The count the number of solutions, we note that there is aone to one correspondence between colorings and a string consisting of N 0's and one 1. The 0's to the left of the 1 represent to objects with color 1 the 0's to the right represent the objects with color 2. The total number of such strings is equal to $\binom{N+1}{1} = N+1$. This means that the coefficient of $x^N$ in $\frac{1}{(1-x)^2}$ is equal to $N+1$.
A: Let $X \sim G(1-x)$, a geometric random variable with success probability $1-x$. We have
$$
\mathbb{E}[X] = \sum_{n=1}^\infty nx^{n-1}(1-x).
$$
On the other hand, we know that $\mathbb{E}[X] = 1/(1-x)$, and we deduce the formula.
We can argue that $\mathbb{E}[X] = 1/(1-x)$ in many ways. One way is to consider $N$ different trials with success probability $1-x$. The number of successful trials is roughly $(1-x)N$, and so the average distance between successful ones (which is distributed according to $X$) is roughly $N/((1-x)N) = 1/(1-x)$. (This argument can be formalized.)
A: As I'd  suggested like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
Using Generalized Binomial Expansion, $$(1+y)^n=1+ny+\frac{n(n-1)}{2!}y^2+\frac{n(n-1)(n-2)}{3!}y^3+\cdots$$ given the converge holds
Comparing with given Series
$ny=2x\  \ \ \ (1)$
$\dfrac{n(n-1)}{2!}y^2=3x^2\  \ \ \ (2)$
$(1)\implies y=\dfrac{2x}n$
From $(2),3x^2=\dfrac{n(n-1)}2\left(\dfrac{2x}n\right)^2\iff n=-2$ as $x\ne0$ for non-trivial cases
$(1)\implies y=\dfrac{2x}n=\dfrac{2x}{-2}=-x$
A: Here is another variation.
Assuming the geometric series $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ is known, we consider functions $f,g:(-1,1) \rightarrow \mathbb{R}$
\begin{align*}
f(x)&=\sum_{n=0}^{\infty}(n+1)x^n\qquad\qquad g(x)=\frac{1}{(1-x)^2}
\end{align*}

We obtain
  \begin{align*}
\int f(x) dx&=\int\left(\sum_{n=0}^{\infty}(n+1)x^n\right) dx
=\sum_{n=0}^{\infty}\int(n+1)x^n dx\\
&=\sum_{n=0}^{\infty}x^{n+1}+C=\sum_{n=1}^\infty x^n+C\\
&=\frac{1}{1-x}-1+C\\
\\
\int g(x) dx&= \int \frac{1}{(1-x)²}dx=\frac{1}{1-x}+D
\end{align*}
We observe $f$ and $g$ have the same antiderivative $\frac{1}{1-x}$ differing by a constant only. Since $f(0)=g(0)=1$ they are equal.

A: $${1\over(1-x)^2}={1\over 1-x}\cdot{1\over 1-x}=\sum_{j\geq0} x^j\cdot\sum_{k\geq0}x^k
=\sum_{r\geq0} x^r\left(\sum_{j+k=r}1\right)=\sum_{r\geq0}(r+1)x^r\ .$$
A: 
ie. $(1 - x)^2(1 + 2x + 3x^2 + ...) = 1$
A: The following proof is far to complicated, but it is a new one and I think it is somewhat funny too.
For $x\in\left[0;1\right)$ we have:
$$
\frac{1}{\left(1-x\right)^2}=\frac{1}{1-\left(2x-x^2\right)}=\sum_{k=0}^{\infty}\left(2x-x^2\right)^k=\sum_{k=0}^{\infty}{\sum_{r=0}^{k}\binom{k}{r}(-1)^r2^{k-r}x^{k+r}}=\sum_{k=0}^{\infty}c_kx^k
$$
We have $c_{2n}=\sum_{s=0}^{n}{\binom{n+s}{n-s}(-1)^{n-s}2^{2s}}$ and $c_{2n+1}=\sum_{s=0}^{n}{\binom{n+s+1}{n-s}(-1)^{n-s}2^{2s+1}}$. Applying the identity $\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}$ we obtain:
$$
c_{2n+2}=\sum_{s=0}^{n+1}{\binom{n+1+s}{n+1-s}(-1)^{n+1-s}2^{2s}}=\sum_{s=0}^{n+1}{\binom{n+s}{n-s}(-1)^{n+1-s}2^{2s}}+\sum_{s=0}^{n+1}{\binom{n+s}{n-s+1}(-1)^{n+1-s}2^{2s}}=-c_{2n}+2c_{2n+1}
$$
$$
c_{2n+3}=\sum_{s=0}^{n+1}{\binom{n+s+2}{n-s+1}(-1)^{n+1-s}2^{2s+1}}=\sum_{s=0}^{n+1}{\binom{n+s+1}{n-s}(-1)^{n+1-s}2^{2s+1}}+\sum_{s=0}^{n+1}{\binom{n+s+1}{n-s+1}(-1)^{n+1-s}2^{2s+1}}=-c_{2n+1}+2c_{2n+2}=3c_{2n+1}-2c_{2n}
$$
Therefore:
$$
c_{2n+3}-c_{2n+2}=3c_{2n+1}-2c_{2n}+c_{2n}-2c_{2n+1}=c_{2n+1}-c_{2n}=…=c_1-c_0=1
$$
Thus:
$$
c_{2n+2}=-c_{2n}+2\left(c_{2n}+1\right)=c_{2n}+2
$$
$$
c_{2n+3}=3c_{2n+1}-2\left(c_{2n+1}-1\right)=c_{2n+1}+2
$$
Together with $c_0=1$ and $c_1=2$ we obtain $c_n=n+1$. Thus, for $x\in\left[0;1\right)$:
$$
\frac{1}{\left(1-x\right)^2}=\sum_{k=0}^{\infty}(k+1)x^k
$$
By observing, that:
$$
\sum_{k=0}^{\infty}(k+1)(-x)^k+\sum_{k=0}^{\infty}(k+1)x^k=\sum_{k=0}^{\infty}(4k+2)x^2k=\frac{4}{\left(1-x^2\right)^2}-\frac{2}{1-x^2}
$$
We get the analogous result for $x\in\left(-1;0\right]$
A: Let $$ f(x) = \sum_{n = 1}^{\infty} nx^{n-1}, \quad |x| < 1. $$ Then $$f'(x) = \sum_{n = 2}^{\infty} n(n-1)x^{n-2} = 2 \left( \sum_{n=2}^{\infty} \frac{n(n-1)}2x^{n-2} \right ).$$ Note $\frac{n(n-1)}2 = \dbinom{n}2$ so $$f'(x) = 2 \sum_{n=2}^{\infty} \dbinom{n}2x^{n-2} .$$ Now consider the coefficient of $x^k$ in $\frac{1}{(1-x)^3} = (1+x+x^2+ \cdots )^3$.The coefficient of $x^k$ is the number of ways to solve the equation $a+b+c = k$ where $0 \le a,b,c \le k$. Imagine this as $k$ dots where we need to place $2$ bars. 
This gives us $\dbinom{k+2}2$ ways. Thus, we have $$ \sum_{n=2}^{\infty} \dbinom{n}2x^{n-2} = \frac{1}{(1-x)^3} $$ so $$f'(x) = \frac{2}{(1-x)^3}$$ so $$f(x) = \int \frac{2}{(1-x)^3} dx = \frac{1}{(1-x)^2} +C. $$ Letting $x = 0$ gives $C = 0$ as desired. 
A: Use differences of a sum.

$$
\begin{array}{r}
S &=& +1 & +2 x & +3 x^2 & +4 x^3 & +5 x^4 & +6 x^5 & +7 x^6 & \cdots\\
-2 x S &=& & -2x & -4 x & -6 x^2 & -8 x^3 & -10 x^4 & -12 x^5 & \cdots\\
x^2 S &=& & & +x^2 & +2 x^2 & +3 x^4 & +4 x^5 & +5 x^6 & \cdots\\
&&&&&&&&&& +\\
\hline\\
\big( 1 - 2 x + x^2 \big) S &=& +1
\end{array}
$$

Or

$$
\begin{array}{rclc}
S &=& \displaystyle + \sum_{k=1}^\infty k x^{k-1}\\
- 2 x S &=& \displaystyle - \sum_{k=1}^\infty 2 k x^{k}\\
x^2 S &=& \displaystyle + \sum_{k=1}^\infty k x^{k+1}\\
&&&+\\
\hline\\
\big( 1 - 2 x + x^2 \big) S
&=& \displaystyle \sum_{k=1}^\infty k x^{k-1}
  - \sum_{k=1}^\infty 2 k x^{k}
  + \sum_{k=1}^\infty k x^{k+1}\\
&=& \displaystyle \sum_{k=0}^\infty (k+1) x^k
  - \sum_{k=1}^\infty 2 k x^{k}
  + \sum_{k=2}^\infty (k-1) x^k\\
&=& \displaystyle 1 + \big[ 2 - 2 \big] x
  + \sum_{k=2}^\infty \big[ (k+1) - 2 k + (k-1) \big] x^k\\
&=& 1
\end{array}
$$

So

$$
S = \frac{1}{1-2x+x^2} = \frac{1}{(1-x)^2}
$$

A: The sequence $y=(1,2,3,4,\ldots)$ is an output of the linear system
$$
y_{k+1}=y_k+u_k,\qquad y_0=1
$$
for the input $u=(1,1,1,1,\ldots)$. Perform the $Z$-transform (multiply by $z^k$ and add up for all $k$)
$$
\sum_{k=0}^\infty y_{k+1}z^k=\underbrace{\sum_{k=0}^\infty y_kz^k}_{y(z)}+\underbrace{\sum_{k=0}^\infty u_kz^k}_{u(z)}\quad\Rightarrow\quad
\frac{1}{z}(y(z)-y_0)=y(z)+u(z)\quad\Rightarrow
$$
$$
\Rightarrow\quad y(z)=\frac{1+z u(z)}{1-z}=\frac{1+z\frac{1}{1-z}}{1-z}=\frac{1}{(1-z)^2}.
$$
A: For the special case $x=\dfrac12$:

If you accept that $1+x+x^2+\dotsb=\dfrac1{1-x}$, the same picture works — just move the horizontal and vertical lines. Instead of them being at $1,1\frac12,1\frac34,\dotsb,2$, you should put them at $1,1+x,1+x+x^2,\dotsb,\dfrac1{1-x}$. The area of the square is then $\left(\dfrac1{1-x}\right){}^2$.

A: Consider $$(1+x+x^2+x^3+\cdots)(1+x+^2+x^3+\cdots)$$ Coefficient of $x^k$ in this expansion is just $k+1.$ Because $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Therefore $$(1+x+x^2+x^3+\cdots)^2=1+2x+3x^2+4x^3+\cdots.$$ 
Now evaluate (you can use the same procedure) $$(1-2x+x^2)(1+2x+3x^2+4x^3+\cdots)=?$$
A: Not a visual proof, but by the Binomial Theorem, $$(1-x)^{-2}=\sum_0^{\infty}{-2\choose n}(-1)^nx^n$$ Now $${-2\choose n}={-2\cdot-3\cdots(-1-n)\over n!}=(-1)^n(n+1)$$ so $(1-x)^{-2}=\sum_0^{\infty}(n+1)x^n$, as desired. 
A: Let $S=1+2x+3x^2+4x^3+\dotsb$.
\begin{align}
\phantom{-x^2}S&=1+2x+3x^2+4x^3+\dotsb\\
\phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\
\phantom{^2}-xS&=\phantom1-\phantom2x-2x^2-3x^3-\dotsb\\
\phantom{-}x^2S&=\phantom{1+2x+2}x^2+2x^3+\dotsb
\end{align}
Adding them together:
\begin{align}
(1-2x&+x^2)S\\
&=1+0x+0x^2+0x^3+\dotsb\\
&=1\\
S&=\frac1{1-2x+x^2}
\end{align}
