Sum of alternating binomial-coefficient-type series

Question

Let $D,n\in \mathbb N$ with $0<D<n$, and $y>0$ is a real number.

Question: Is there a closed-form for the following alternating sequence \begin{equation} \sum_{k=0}^D (-y)^k {n\choose k}? \end{equation}

This question is related to a similar question where $y=1$.

Motivation: These kind of expressions arrive in numerical schemes that I am working on.

First attempt: I tried to repeat some of the techniques from $y=1$ case. For instance, using the complex integral argument, we have \begin{align} \sum_{k=0}^D (-y)^k {n\choose k} &= \sum_{k=0}^D (-y)^k \oint_{0<|z|<1} \frac{(1+z)^n}{z^{k+1}}\frac{dz}{2\pi i} = \oint_{0<|z|<1} \frac{(1+z)^n}{z} \sum_{k=0}^D \left(-\frac{y}{z}\right)^k \frac{dz}{2\pi i} \\ &= \oint_{0<|z|<1} \frac{(1+z)^n}{z} \left[\frac{z+y(-y/z)^D}{z+y}\right] \frac{dz}{2\pi i} \\ &= \oint_{0<|z|<1} \frac{(1+z)^n}{z+y} \frac{dz}{2\pi i} + y(-y)^D \oint_{0<|z|<1} \frac{(1+z)^n}{z^{D+1}(z+y)} \frac{dz}{2\pi i}. \end{align} I don't know much about complex integrals and so don't know how to complete this argument. Any ideas?

Or are their any other techniques that will work here? Any ideas/hints are appreciated.

Answer 1

We show the difference between the variant with general $y$ and $y=1$. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k}=\oint_{0<|z|<1} \frac{(1+z)^n}{z^{k+1}}\frac{dz}{2\pi i} \tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^D(-y)^k\binom{n}{k}} &=\sum_{k=0}^D[z^k](1-yz)^n\tag{2}\\ &=[z^0](1-yz)^n\sum_{k=0}^Dz^{-k}\tag{3}\\ &=[z^0](1-yz)^n\frac{z^{-(D+1)}-1}{z^{-1}-1}\\ &=[z^{-1}](1-yz)^n\frac{z^{-(D+1)}-1}{1-z}\\ &=\left([z^D]-[z^{-1}]\right)\frac{(1-yz)^n}{1-z}\tag{4}\\ &\,\,\color{blue}{=[z^D]\frac{(1-yz)^n}{1-z}}\tag{5} \end{align*}

We see in (5) that in case $y=1$ we get \begin{align*} [z^D]\left.\frac{(1-yz)^n}{1-z}\right|_{y=1}=[z^D](1-z)^{n-1}=(-1)^D\binom{n-1}{D} \end{align*} but there is no way to make similar simplifications with general $y$.

Comment:

• In (2) we apply the coefficient of operator according to (1).

• In (3) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

• In (4) we note $\frac{(1-yz)^n}{1-z}=(1-yz)^n\sum_{j=0}^\infty z^j$ is a power series in $z$ so that the residue $[z^{-1}]$ is zero.

Answer 2

You could write it using a hypergeometric function:

$$ \left( 1-y \right) ^{n}-{n\choose D+1} \left( -y \right) ^{D+1} {\mbox{$_2$F$_1$}(1,-n+D+1;\,D+2;\,y)} $$