Mass function; negative binomial distribution

Question

I want to prove that the mass function of the negative binomial distribution satisfies that it adds up to $1\;\forall x\in\mathbb R$.

My textbook starts as follows: $\displaystyle\sum_{k=n}^{\infty}\left(\begin{array}{l}k-1 \\ n-1\end{array}\right) p^{n} q^{k-n}=p^{n} \sum_{l=0}^{\infty}\left(\begin{array}{c}n+l-1 \\ l\end{array}\right) q^{l}=p^{n} \sum_{l=0}^{\infty}\left(\begin{array}{c}-n \\ l\end{array}\right)(-q)^{l}$

I don't understand how they substituted $l$ in ${k-1\choose n-1}$. I would have done the following:

$p^{n} \sum_{l=0}^{\infty}\left(\begin{array}{l}n+l-1 \\ k-l-1\end{array}\right) q^{l}$ - but this way I don't get their result.

Could someone help me with this?

(I am new here, so I'm not allowed to post pictures yet. That's why I'm posting links instead of pictures).

Answer 1

Binomial coefficients are symmetric. \begin{eqnarray*} \binom{ k-1}{ n-1}=\binom{ k-1}{ k-n} \end{eqnarray*} Does this help ? Sha, Highlight the above equation & right click, then choose the option show math as & then TeX command ... it will show you the tex commands to write equations.