Partial sum of binomial expansion

Question

Let $0<p<1$ and $n>m>0.$ I am looking for a closed form of

$\sum_{k=0}^{m}{n-k\choose m-k}p^k$ or $\sum_{k=0}^{m}{n-m+k\choose k}p^{m-k}$

Answer 1

An answer is:$\frac{\left(n-m+1\right)_{m}}{m!}\cdot_{2}F_{1}\left(\begin{array}{c} -m,1\\-n\end{array};p\right)$
Converting partial sums to Generalized Hypergeometric functions may just seem like repackaging the calculations but there is a wonderful side-effect. For instance, in this application one gains a multitude of identities. These are the formulas from section 15.5 of the DLMF http://dlmf.nist.gov/15.5 . The contiguous identities are straightforward but the derivatives can be used as is or, since we know the derivative explicit forms, expanded to give other identities, x+y=z. More complicated partial sums of binomials can be derived from section 16.3 of the DLMF http://dlmf.nist.gov/16.3.
Tutorial on converting combinatorial partial sums to closed form
Unfortunately, I am slightly dyslexic so somebody should check the following calculations.
Prerequisites:
$n!=\Gamma\left(n+1\right),\left(a\right)_{n}=\frac{\Gamma\left(a+n\right)}{\Gamma\left(a\right)}$
$_{p}F_{q}\left(\begin{array}{c} \left(a_{j}\right)_{k}\\ \left(b_{i}\right)_{k} \end{array};p\right)={\displaystyle \sum_{k=0}^{\infty}\frac{{\displaystyle \prod_{j=0}^{p}\left(a_{j}\right)_{k}}}{{\displaystyle \prod_{i=0}^{q}}\left(a_{j}\right)_{k}}\cdot\frac{p^{k}}{k!}}$
$p^{k}=\frac{\Gamma\left(k+1\right)}{k!}p^{k}=\frac{\left(1\right)_{k}}{k!}p^{k}$

Where the $_{p}F_{q}\left(\right)$ looks hard but really isn't as I shall show.
In order to use the formulas above easily, all of the summation indices's $k$ should have a positive sign. This may require some manipulation of the binomial in general; but not in the second part of this particular problem.

1. We convert the binomial to $\Gamma\left(\right)$ and then $\left(a_{j}\right)_{k}$

$\left(\begin{array}{c} n-m+k\\ k \end{array}\right)=\frac{\Gamma\left(n-m+1+k\right)}{\Gamma\left(1+k\right)\cdot\Gamma\left(n-m+1\right)}=\left(\frac{\Gamma\left(n-m+1+k\right)}{\Gamma\left(n-m+1\right)}\right)\cdot\frac{1}{\left(\frac{\Gamma\left(1+k\right)}{\Gamma\left(1\right)}\right)}=\frac{\left(n-m+1\right)_{k}}{\left(1\right)_{k}}$

2. Sum and form the hypergeometric representation

$p^{m}\cdot{\displaystyle \sum_{k=0}^{\infty}}\frac{\left(n-m+1\right)_{k}}{\left(1\right)_{k}}\cdot\frac{\left(1\right)_{k}}{1}\frac{p^{-k}}{k!}=p^{m}\,_{1}F_{0}\left(\begin{array}{c} n-m+1\\ - \end{array};\frac{1}{p}\right)$
Where the dash stands for 1 and the $p^{m}$ has to remain tied to the hypergeometric term. Otherwise the calculation for $_{1}F_{0}$ will diverge at zero. This is resolved in the next step but be warned in general; this is resolved in DLMF 16.2.4

3. Restrict the summation to m terms by DLMF 16.2.4 : http://dlmf.nist.gov/16.2.E4

$p^{m}\cdot\frac{\left(n-m+1\right)_{m}}{1}\cdot\frac{p^{-m}}{m!}\cdot_{2}F_{1}\left(\begin{array}{c} -m,1\\ 1-m-(n-m+1) \end{array};\left(-1\right)^{1+0+1}p\right)$
If you are worried about the lack of a $(b)$ term you can just add a dummy $\left(c\right)$ to the numerator and denominator and keep it to cancel out after the conversion.

4. Resolve some terms for

$\frac{\left(n-m+1\right)_{m}}{m!}\cdot_{2}F_{1}\left(\begin{array}{c} -m,1\\ -n \end{array};p\right)$

Notice the $-n$ in the bottom; generally this is a sign of trouble. But this is balanced by the $-m$ on the top so the stepping $-m,\left(-m+1\right),\left(-m+2\right)$ will reach zero before $-n$ does.

Now we have a perfectly good analytic function that can be manipulated.

Constructive comments welcome!