Truncated alternating binomial sum It is easily checked that
$\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0$, for example by appealing to the binomial theorem.
I'm trying to figure out what happens with the truncated sum
$\displaystyle\sum_{i\ =\ 0}^{D}\left(\, -1\,\right)^{i}\binom{n}{i}$.

How far away from $0$ can this get, as a function of $D$ ?.

I'm mostly interested in the case of when $D \ll n$, such as
$D \sim \,\sqrt{\,n\,}\,$.
Thanks !
 A: Hint: The answer will be $(-1)^D{n-1\choose D}$. The proof goes by induction on $D$, and uses the Pascal triangle rule.

This answer was moved here from an older question due to a merger. In that question the parameter $D$ was called $k$. I edited this answer to match with that.
A: First, let us write $$\sum_{i=0}^{D} (-1)^i \binom{n}{i}=\sum_{i=0}^D \binom{i-n-1}{i}$$
This step can be proven by using the definition of binomial coefficient and pulling a $-1$ out of each term.
Next, $$\sum_{i=0}^D \binom{i-n-1}{i}=\binom{D-n}{D}$$ can be proven inductively.
And finally, this can be simplified using the same result in the first step.
$$\binom{D-n}{D}=(-1)^D\binom{n-1}{D}$$
A: Use the remainder term of a truncated Taylor series:
$$(1+x)^n=\sum_{k=0}^n{n \choose k}x^k=\sum_{k=0}^r{n \choose k}x^k + \int_0^x \frac{f^{(r+1)}(t)}{r!}(x-t)^r\,dt$$
Rearrange to put the truncated series on the LHS:
$$\begin{align}
\sum_{k=0}^r{n \choose k}x^k &= (1+x)^n - \int_0^x \frac{f^{(r+1)}(t)}{r!}(x-t)^r\,dt 
\\&= (1+x)^n - \frac{1}{r!}\frac{n!}{(n-r-1)!} \int_0^x (1+t)^{n-r-1}(x-t)^r\,dt
\end{align}$$
Substitute $x=-1$:
$$\begin{align}
\sum_{k=0}^r{n \choose k}(-1)^k &=  0-\frac{1}{r!}\frac{n!}{(n-r-1)!} (-1)^{r}\int_0^{-1} (1+t)^{n-1}\,dt\\
&=\frac{(-1)^{r+1}}{r!}\frac{(n-1)!}{(n-r-1)!}[(1+t)^n]_0^{-1}
\\&=\frac{(-1)^{r}}{r!}\frac{(n-1)!}{(n-r-1)!}
\\&=(-1)^r{n-1\choose r}
\end{align}$$
A: Bijective proof
In this answer, I'm assuming the identity is already known, but a bijective proof is sought.
Rearrange $$\sum_{i=0}^{D}(-1)^i\binom{n}{i}=(-1)^D\binom{n-1}{D}$$
to get the equivalent statement $$\sum_{i=0}^{D}\binom{n}{i}[i\text{ even}] + [D\text{ odd}]\binom{n-1}{D} = \sum_{i=0}^{D}\binom{n}{i}[i\text{ odd}] + [D\text{ even}]\binom{n-1}{D}$$
where I'm using Iverson Bracket notation.
Let $2^{[n]}$ denote the set of strings of length $n$ made up of $0$s and $1$s. Consider a transformation $\mathtt{FLIPFIRST}:2^{[n]} \to 2^{[n]}$ that flips the first digit. A bijective proof of the above claim can be found using $\mathtt{FLIPFIRST}$.
Observe that $\mathtt{FLIPFIRST}$ is injective, because $\mathtt{FLIPFIRST}(\mathtt{FLIPFIRST}(s))=s$. $\mathtt{FLIPFIRST}$ maps a string with an odd number of $1$ to one with an even number of $1$s, and vice versa. If a string has $D$ $1$s, then its output may have $D+1$ or $D-1$ $1$s. The number of strings $s$ such that $\mathtt{FLIPFIRST}(s)$ has $D+1$ $1$s is $\binom{n-1}{D}$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$\begin{align}
&\color{#88f}{\large\sum_{k = 0}^{D}\pars{-1}^{k}{n \choose k}}
=\sum_{k = 0}^{D}\pars{-1}^{k}\ \overbrace{\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}
{\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ {n \choose k}}}
\\[5mm]& \
=\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{n} \over z}
\sum_{k = 0}^{D}\pars{-\,{1 \over z}}^{k}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\ \oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}{\pars{1 + z}^{n} \over z}
{\pars{-1/z}^{D} + z \over 1 + z}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\pars{-1}^{D}\ \underbrace{\oint_{\verts{z}\ =\ a\ <\ 1}
{\pars{1 + z}^{n - 1} \over z^{D + 1}}\,{\dd z \over 2\pi\ic}}
_{\ds{=\ {n - 1 \choose D}}}\ +\
\underbrace{\oint_{0\ <\ \verts{z}\ =\ a\ <\ 1}\pars{1 + z}^{n - 1}
\,{\dd z \over 2\pi\ic}}_{\ds{=\ 0}}
\\[5mm]&\ =
\bbox[10px,border:1px groove navy]{\pars{-1}^{D}{n - 1 \choose D}}
\\ &
\end{align}
A: Let $n\ge 2\in\mathbb N$ (since the case $n=1$ is trivial). 
For $0\le D\lt n$, we can prove the following by induction:
$$\sum_{i=0}^{D}(-1)^i\binom{n}{i}=(-1)^D\binom{n-1}{D}.$$
For $D=0$, it holds trivially.
For a $D$ such that $0\le D\le n-2$, suppose that it holds. Then,
$$\begin{align}\sum_{i=0}^{D+1}(-1)^i\binom{n}{i}&=(-1)^{D+1}\binom{n}{D+1}+\sum_{i=0}^{D}(-1)^i\binom{n}{i}\\&=(-1)^{D+1}\binom{n}{D+1}+(-1)^D\binom{n-1}{D}\\&=(-1)^{D+1}\left\{\binom{n}{D+1}-\binom{n-1}{D}\right\}\\&=(-1)^{D+1}\binom{n-1}{D+1}\end{align}$$
Hence, it holds when $D+1$.
Therefore, it holds for any $0\le D\lt n$. Q.E.D.
From this, you'll also see how far away from $0$ it can get. 
A: Use the following:
$$ (1-x)^{n-1} = (1-x)^n \times \frac{1}{1-x} = (1-x)^n (1 + x + x^2 + \dots) =$$
$$\left(1 + n(-x) + \binom{n}{2}(-x)^2 + \dots + (-x)^n\right)(1+x+x^2 + \dots) $$
Now, mutiplying any polynomial (or power series) by $1 + x + x^2 + \dots$ has the effect of giving you the truncated sums of the coefficients of the polynomial as the coefficients of the powers of $x$ in the resulting power series.
In your case, the resulting series is itself a polynomial, $(1-x)^{n-1}$, giving you a neat closed form answer.
A: Let us write $$S_{n,m}=\sum_{k=o}^{m} (-1)^k~ {n \choose k},$$  then
$$S_{n,m}={n \choose 0}-{n \choose 1}+{n\choose 2}-{n\choose 3}+......+ (-1)^m {n \choose m}.$$
$$ \Rightarrow S_{n,m}=
[x^0]~ \left ((1-x)^n+(1-x)^n \frac{1}{x}+(1-x)^n \frac{1}{x^2}+(1-x)^n\frac{1}{x^3}+....+(1-x)^n \frac{1}{x^m}\right).$$
$[x^j]~$ is short for `coefficient of $x^j$ in '. We can write
$$S_{n,m}=[x^0] ~ (1-x)^n \left(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+...+\frac{1}{x^m}\right).$$
$$\Rightarrow S_{n,m}=[x^0]~(1-x)^n \frac{1}{x^m} \left(\frac{1-x^{m+1}}{1-x} \right)= [x^m]~[ (1-x)^{n-1}-(1-x)^{n-1}  x^{m+1}].$$
$$\Rightarrow S_{n,m}= (-1)^m {n-1 \choose m}$$ 
A: Just to show another way
$$
\eqalign{
  & \sum\limits_{i = 0}^D {\left( { - 1} \right)^{\,i} \left( \matrix{  n \cr   i \cr}  \right)}
  = \sum\limits_{i = 0}^D {\left( \matrix{  i - n - 1 \cr   i \cr}  \right)}  =   \cr 
  &  = \sum\limits_i {\left( \matrix{  D - i \cr   D - i \cr}  \right)
\left( \matrix{  i - n - 1 \cr   i \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,D} \sum\limits_i {\left( \matrix{   - 1 \cr   D - i \cr}  \right)
\left( \matrix{  n \cr   i \cr}  \right)}  =   \cr 
  &  = \left( { - 1} \right)^{\,D} \left( \matrix{  n - 1 \cr   D \cr}  \right)
\quad \left| \matrix{
  \;n \in \mathbb C \hfill \cr 
  \;0 \le D \in \mathbb Z \hfill \cr}  \right.
 \cr} 
$$
where the steps are:

*

*upper negation;

*replacing the bounds on the sum with a binomial;

*upper negation on both binomials;

*Vandermonde convolution.

A: With @Aryabhata:'s idea:
Consider the binomial expansion
$$(1+x)^{\alpha} = \sum_{k\ge 0} \binom{\alpha}{k} x^k$$
where
$$\binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$$
and $0$ for  integers $k<0$.
Since
$$(*) \ \ \ \ (1+x)^{\alpha}\cdot (1+x)^{\beta} = (1+x)^{\alpha+ \beta}$$
we get
$$(**) \ \ \ \sum_{l \ge 0} \binom{\alpha}{k-l} \binom{\beta}{l} = \binom{\alpha+ \beta}{k}$$
The above is true for any $\alpha$, $\beta$, and $k\ge 0$ an integer.
To get the desired formula, notice that $\binom{-1}{l} = (-1)^l$.
Notes:

*

*The binomial coefficient $\binom{\alpha}{k}$ is can be defined for all $\alpha$ and $k$ ( complex).  The  formula $(**)$ is still valid for $k$ non-integer (but now summing infinitely many terms).


*The binomial identity $(**)$ is equivalent to the equality $(*)$. It can be also proves as follows: consider it as an polynomial for variables $\alpha$, $\beta$. Since it is true for any pair $(\alpha, \beta) = (m,n)$, it is an identity.


*We can modify the binomial coefficients to
$$\binom{\alpha}{k}_h\colon = \frac{\alpha(\alpha-h)\cdots(\alpha - (k-1) h)}{k!}$$
Now the equality
$$\binom{\alpha+ \beta}{r}_h = \sum_{p+q=r} \binom{\alpha}{p}_h \cdot \binom{\beta}{q}_h$$
as $h\to 0$ becomes the binomial formula for $(\alpha+\beta)^r$.
