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 !

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.
