mean and variance formula for negative binomial distribution The equation below indicates expected value of negative binomial distribution. I need a derivation for this formula. I have searched a lot but can't find any solution. Thanks for helping :)
$$
E(X)=\sum_{x=r}^\infty x\cdot \binom {x-1}{r-1} \cdot p^r \cdot (1-p)^{x-r} =\frac{r}{p}
$$
I have tried:
\begin{align}
E(X) & =\sum _{x=r} x\cdot \binom{x-1}{r-1} \cdot p^r \cdot (1-p)^{x-r} \\[8pt]
& = \sum_{x=r}^\infty x \cdot \frac{(x-1)!}{(r-1)! \cdot ((x-1-(r-1))!} \cdot p^r \cdot (1-p)^{x-r} \\[8pt]
& = \sum_{x=r}^\infty \frac{x!}{(r-1)!\cdot ((x-r)!} \cdot p^r \cdot (1-p)^{x-r} \\[8pt]
\Longrightarrow & \phantom{={}} \sum_{x=r}^\infty r\cdot \frac{x!}{r!\cdot (x-r)!}\cdot p^r \cdot (1-p{)}^{x-r} \\[8pt]
& = \frac{r}{p} \cdot \sum_{x=r}^\infty \frac{x!}{r!\cdot (x-r)!}\cdot p^{r+1}\cdot (1-p)^{x-r}
\end{align}
If the power of $p$ in the last equation were not $r + 1,$ I can implement Newton Binomial. So It will be true. But I am stuck here.

Here is an alternative derivation to compute the expectation of a negative binomial distribution. Let $X_r$ be the number of trials needed to get $r$ successes, and $p$ be the probability of success on any given trial. If $r=1$, then $X_r$ has a geometric distribution, and so $\mathrm{E}(X_r)=1/p$. To prove this, let $S$ be the event that the first trial is success. By the law of iterated expectation,
\begin{align}
\DeclareMathOperator{\E}{\mathrm{E}}
\DeclareMathOperator{\P}{\mathrm{P}}
\E(X_1) &= \E(X_1 \mid S)\P(S)+\E(X_1\mid S')P(S') \\[4pt]
&= (1 \cdot p)+(\E(X_1)+1)(1-p) \, ,
\end{align}
and rearranging yields the desired result. Computing $\E(X_r)$ in general is not too difficult if we use linearity of expectation. Let $A_i$ be the number of additional trials needed to get the $i$-th success once you have had $i-1$ successes. Note that $X_r=\sum_{i=1}^{r}A_i$, and so
$$
\E(X_r)=\E\left(\sum_{i=1}^r A_i\right)=\sum_{i=1}^r \E(A_i) \, .
$$
But the expected number of trials needed to get the $i$-th success is no different to the expected number of trials needed to get the first success, and so $\E(A_i)=\E(A_1)=\E(X_1)=1/p$. Hence, $E(X_r)=r/p$.
