No need to index twice. The process is quite similar to the way you get $E[x]$

I believe the problem here is how to get $E[x^2]$

Please refer to Markus Scheuer's efforts:

In his answer he derived: $$ \begin{align*} E(x)&=rp^r\sum_{k=0}^{\infty}\binom{k+r}{k}(1-p)^k\\ \end{align*}. $$ Follow the similar way, and apply to $E[X^2]$ $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ \end{align*} $$ We need a way to change (k+r) to (k+r-1), then the outside (k+r) can be combined with (k+r-1)! as (k+r)!, so I refer to the wiki page $$ \begin{align*} \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}\\ \end{align*} $$

$$ \begin{align*} (k+r)\binom{k+r}{k}&=(k+r)\binom{k+r-1}{k-1}+(k+r)\binom{k+r-1}{k}\\ &=(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}\\ \end{align*} $$ So: $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}](1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}](1-p)^k+rp^r\sum_{k=0}^{\infty}[r\binom{k+r}{k}](1-p)^k\\ \end{align*} $$ then, still, use Markus Scheuer's idea in remember to use binomial series expansion

Finally, you will get $$ \begin{align*} E(x^2)&=\frac{r(1-r-p)}{p^2}\\ \end{align*} $$

$$ \begin{align*} V(X)&=E(x^2)-[E(x)]^2\\ &=\frac{r(1-r-p)}{p^2}+\frac{r^2}{p^2}\\ &=\frac{r(1-p)}{p^2} \end{align*} $$

No need to index twice. The process is quite similar to the way you get $E[x]$

I believe the problem here is how to get $E[x^2]$

Please refer to Markus Scheuer's efforts:

In his answer he derived: $$ \begin{align*} E(x)&=rp^r\sum_{k=0}^{\infty}\binom{k+r}{k}(1-p)^k\\ \end{align*}. $$ Follow the similar way, and apply to $E[X^2]$ $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ \end{align*} $$ We need a way to change $(k+r)$ to $(k+r-1)$, then the outside $(k+r)$ can be combined with $(k+r-1)!$ as $(k+r)!$, so I refer to the wiki page $$ \begin{align*} \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}\\ \end{align*} $$

$$ \begin{align*} (k+r)\binom{k+r}{k}&=(k+r)\binom{k+r-1}{k-1}+(k+r)\binom{k+r-1}{k}\\ &=(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}\\ \end{align*} $$ So: $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}](1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}](1-p)^k+rp^r\sum_{k=0}^{\infty}[r\binom{k+r}{k}](1-p)^k\\ \end{align*} $$ then, still, use Markus Scheuer's idea in remember to use binomial series expansion

Finally, you will get $$ \begin{align*} E(x^2)&=\frac{r(1-r-p)}{p^2}\\ \end{align*} $$

$$ \begin{align*} V(X)&=E(x^2)-[E(x)]^2\\ &=\frac{r(1-r-p)}{p^2}+\frac{r^2}{p^2}\\ &=\frac{r(1-p)}{p^2} \end{align*} $$

Hello~ I find a way to get the variance.

No need to index twice. The process is quite similar to the way you get $E[x]$

I believe the problem here is how to get $E[x^2]$

Please refer to Markus Scheuer's efforts: Proof for the calculation of mean in negative binomial distribution

In his answer he derived: $$ \begin{align*} E(x)&=rp^r\sum_{k=0}^{\infty}\binom{k+r}{k}(1-p)^k\\ \end{align*} $$

Follow the similar way, and apply to $E[X^2]$

 $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ \end{align*} $$

This (k+r) is so annoying! Because in the binomial function, there is another (k+r)

We need a way to change (k+r) to (k+r-1), then the outside (k+r) can be combined with (k+r-1)! as (k+r)!

  so I refer to:https://en.wikipedia.org/wiki/Binomial_coefficient $$ \begin{align*} \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}\\ \end{align*} $$

$$ \begin{align*} (k+r)\binom{k+r}{k}&=(k+r)\binom{k+r-1}{k-1}+(k+r)\binom{k+r-1}{k}\\ &=(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}\\ \end{align*} $$

So:

 $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}](1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}](1-p)^k+rp^r\sum_{k=0}^{\infty}[r\binom{k+r}{k}](1-p)^k\\ \end{align*} $$

then, still, use Markus Scheuer's idea in Proof for the calculation of mean in negative binomial distribution

remember to use binomial series expansion

Finally, you will get $$ \begin{align*} E(x^2)&=\frac{r(1-r-p)}{p^2}\\ \end{align*} $$

$$ \begin{align*} V(X)&=E(x^2)-[E(x)]^2\\ &=\frac{r(1-r-p)}{p^2}+\frac{r^2}{p^2}\\ &=\frac{r(1-p)}{p^2} \end{align*} $$

emmm, this is my first time to answer a question. this is also my first time to use Latax... Sorry that I don't know how to make beautiful format. Hope this can help you and others.

No need to index twice. The process is quite similar to the way you get $E[x]$

I believe the problem here is how to get $E[x^2]$

Please refer to Markus Scheuer's efforts:

In his answer he derived: $$ \begin{align*} E(x)&=rp^r\sum_{k=0}^{\infty}\binom{k+r}{k}(1-p)^k\\ \end{align*}. $$ Follow the similar way, and apply to $E[X^2]$  $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ \end{align*} $$ We need a way to change (k+r) to (k+r-1), then the outside (k+r) can be combined with (k+r-1)! as (k+r)!, so I refer to the wiki page $$ \begin{align*} \binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}\\ \end{align*} $$

$$ \begin{align*} (k+r)\binom{k+r}{k}&=(k+r)\binom{k+r-1}{k-1}+(k+r)\binom{k+r-1}{k}\\ &=(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}\\ \end{align*} $$ So:  $$ \begin{align*} E(x^2)&=rp^r\sum_{k=0}^{\infty}(k+r)\binom{k+r}{k}(1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}+r\binom{k+r}{k}](1-p)^k\\ &=rp^r\sum_{k=0}^{\infty}[(r+1)\binom{k+r}{k-1}](1-p)^k+rp^r\sum_{k=0}^{\infty}[r\binom{k+r}{k}](1-p)^k\\ \end{align*} $$ then, still, use Markus Scheuer's idea in remember to use binomial series expansion

Finally, you will get $$ \begin{align*} E(x^2)&=\frac{r(1-r-p)}{p^2}\\ \end{align*} $$

$$ \begin{align*} V(X)&=E(x^2)-[E(x)]^2\\ &=\frac{r(1-r-p)}{p^2}+\frac{r^2}{p^2}\\ &=\frac{r(1-p)}{p^2} \end{align*} $$