$\sum_{n=0}^{\infty} (2n+1)x^n$ Closed Form 
$\sum_{n=0}^{\infty} (2n+1)x^n$ Closed Form

I'm trying to find the closed form for the specified series. However, I'm having a bit of trouble doing so. I assume there's a technique here that I haven't quite figured out, and any help would be appreciated.
Here's what I know: 
$\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ 
And a technique on taking derivatives to find some other closed froms.
I'm not sure if I can apply any of these on the desired series.
 A: $\displaystyle \sum_{n=0}^{\infty}(2n+1)x^{n}$ 
$\displaystyle = \sum_{n=0}^{\infty}[(2n+2)-1]x^{n}$ 
$\displaystyle = 2\sum_{n=0}^{\infty}(n+1)x^{n} - \sum_{n=0}^{\infty}x^{n}$
$\displaystyle = 2\sum_{n=0}^{\infty} \frac{\mathrm{d}}{\mathrm{d}x}x^{n+1} - \frac{1}{1-x}$
$\displaystyle = 2 \frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=0}^{\infty} x^{n+1} - \frac{1}{1-x}$
$\displaystyle = 2 \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x}{1-x} \right) - \frac{1}{1-x}$
$\displaystyle = \frac{2}{(1-x)^2}  - \frac{1}{1-x}$
$\displaystyle = \frac{1+x}{(1-x)^2}$
Note that $|x| < 1$
A: Without calculus.
Start with
$f_0(x)
=\sum_{n=0}^{\infty} x^n
=\dfrac1{1-x}
$.
If
$f_1(x)
=\sum_{n=0}^{\infty} nx^n
$,
then
$f_1(x)-xf_1(x)
=(1-x)f_1(x)
$
and
$\begin{array}\\
f_1(x)-xf_1(x)
&=\sum_{n=0}^{\infty} nx^n-x\sum_{n=0}^{\infty} nx^n\\
&=\sum_{n=0}^{\infty} nx^n-\sum_{n=0}^{\infty} nx^{n+1}\\
&=\sum_{n=0}^{\infty} nx^n-\sum_{n=1}^{\infty} (n-1)x^{n}\\
&=\sum_{n=0}^{\infty} nx^n-\left(\sum_{n=1}^{\infty} nx^{n}-\sum_{n=1}^{\infty} x^{n}\right)\\
&=\sum_{n=1}^{\infty} x^{n}\\
&=f_0(x)-1\\
&=\dfrac1{1-x}-1\\
&=\dfrac{1-(1-x)}{1-x}\\
&=\dfrac{x}{1-x}\\
\end{array}
$
so
$f_1(x)
=\dfrac{x}{(1-x)^2}
$.
Then
$\begin{array}\\
\sum_{n=0}^{\infty}(2n+1)x^n
&=\sum_{n=0}^{\infty}(2n)x^n+\sum_{n=0}^{\infty}x^n\\
&=2\sum_{n=0}^{\infty}nx^n+f_0(x)\\
&=2\dfrac{x}{(1-x)^2}+\dfrac1{1-x}\\
&=\dfrac{2x+(1-x)}{(1-x)^2}\\
&=\dfrac{1+x}{(1-x)^2}\\
\end{array}
$
A: HINT
To begin with, notice that
\begin{align*}
\sum_{n=0}^{\infty}(2n+1)x^{n} = \sum_{n=0}^{\infty}[(2n+2)-1]x^{n} = 2\sum_{n=0}^{\infty}(n+1)x^{n} - \sum_{n=0}^{\infty}x^{n}
\end{align*}
Then you make use of the fact that
\begin{align*}
\begin{cases}
\displaystyle\sum_{n=0}^{\infty}x^{n} = \frac{1}{1-x} & \text{for}\,\,|x| < 1\\\\
\displaystyle (n+1)x^{n} = \frac{\mathrm{d}}{\mathrm{d}x}x^{n+1}
\end{cases}
\end{align*}
A: see Help with summation: $\sum_{k=1}^\infty\frac{k(k+2)}{15^k}$
Reusing that post result: $s_0 = \frac{x}{1-x}, s_1 = \frac{x}{(1-x)^2}$
$$\sum_{n=0}^\infty (2n+1) x^n = 1 + \sum_{n=1}^\infty (2n+1) x^n =
1 + 2 s_1 + s_0 = \frac{1+x}{(1-x)^2}$$
A: Hint
$$\sum_{n=0}^{\infty} (2n+1)x^n=2x\sum_{n=0}^{\infty} nx^{n-1}+\sum_{n=0}^{\infty} x^{n}=2x \left(\sum_{n=0}^{\infty} x^{n} \right)'+\left(\sum_{n=0}^{\infty} x^{n} \right)$$
A: So basically we need to find , $$2 \sum_{n=0}^{\infty}nx^n + \sum_{n=0}^{\infty}x^n $$
First we'll solve for $  \sum_{n=0}^{\infty}x^n$ . Since $|x|<1$
$$ \sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$
Now $\sum_{n=0}^{\infty}nx^n $ will be : $$ \sum_{n=0}^{\infty}nx^n=1x+2x^2+3x^3+.......\infty$$
So we need to solve for $1x+2x^2+3x^3+.......\infty$ this series . Let this series be defined as $\Lambda$
$$\Lambda=1x+2x^2+3x^3+.......\infty$$ $$\frac{\Lambda}{x}=1+2x+3x^2+.......\infty$$
Now , $$\frac{\Lambda}{x}-\Lambda=1+x+x^2+x^3+.....\infty$$ $$\Lambda\left(\frac{1-x}{x}\right)=\frac{1}{1-x}$$ $$\Lambda=\frac{x}{(1-x)^2}$$
FINALLY : $$2 \sum_{n=0}^{\infty}nx^n + \sum_{n=0}^{\infty}x^n =\frac{2x}{(1-x)^2}+\frac{1}{1-x}=\frac{1}{1-x}\left(\frac{2x+1-x}{1-x}\right)=\frac{x+1}{(x-1)^2}$$
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