The sum of the odd terms of partial combination numbers

Question

I have two following sums $ S_1 = \sum_{k=1}^{K} \binom{N}{2k-1}$, $ S_2 = \sum_{k=1}^{K} \binom{N}{2k}$. When $2K=N$, it is easy to obtain the results, however, is there a closed-form solution for $2K<N$? Or just a tight upper/lower bound? Any comments would be appreciated!

Answer 1

We can rewrite the sums of interest as $$ \eqalign{ & S_{\,1} = \sum\limits_{k = 1}^K {\left( \matrix{ N \cr 2\,k - 1 \cr} \right)} = \sum\limits_{k = 0}^{K - 1} {\left( \matrix{ N \cr 2\,k + 1 \cr} \right)} \cr & S_{\,2} = \sum\limits_{k = 1}^K {\left( \matrix{ N \cr 2\,k \cr} \right)} = \sum\limits_{k = 0}^K {\left( \matrix{ N \cr 2\,k \cr} \right)} - 1 = S_{\,0} - 1 \cr} $$ Then consider that $$ \sum\limits_{k = 0}^K {\left( \matrix{ N \cr 2\,k \cr} \right)x^{\,2k} } + \sum\limits_{k = 0}^{K - 1} {\left( \matrix{ N \cr 2\,k + 1 \cr} \right)x^{\,2k + 1} } = \sum\limits_{k = 0}^{2K} {\left( \matrix{ N \cr \,k \cr} \right)x^{\,k} } $$ Therefore we can concentrate on the Truncated Binomial $$ B_{\,T} (z,r,m) = \sum\limits_{k = 0}^m {\left( \matrix{ r \cr k \cr} \right)z^{\,k} } \quad \left| \matrix{ \;r \in R \hfill \cr \;0 \le m \in Z \hfill \cr \;z \in C \hfill \cr} \right. $$ since therefrom we can retrieve the requested sums as $$ \eqalign{ & S_{\,1} = {{B_{\,T} (1,N,2K) - B_{\,T} ( - 1,N,2K)} \over 2} \cr & S_{\,2} = {{B_{\,T} (1,N,2K) + B_{\,T} ( - 1,N,2K)} \over 2} - 1 \cr } $$

The truncated binomial is just a simple polynomial of degree $m$ in $z$: no way to look for a "simpler" expression.
However it might be useful to "render" it in other forms, to show possible algebraic manipulations , specially for asymptotics.
I will concisely resume some alternative representations in the following.

a) Double o.g.f.

We have that taking the o.g.f. both on $k$ and $m$ $$ \bbox[lightyellow] { $$ \bbox[lightyellow] { \eqalign{ & F_{\,a} (x,z,r)\quad \left| \matrix{ \;0 \le m \in \mathbb Z \hfill \cr \;z,x,r \in \mathbb C \hfill \cr} \right.\quad = \cr & = \sum\limits_{0\, \le \,\,m} {\left( {\sum\limits_{\left( {0\, \le } \right)\,k\, \le \,m} {\left( \matrix{ r \cr k \cr} \right)x^{\,k} } } \right)z^{\,m} } = \cr & = \sum\limits_{0\, \le \,\,m} {\sum\limits_{\left( {0\, \le } \right)\,k\, \le \,m} {\left( \matrix{ r \cr k \cr} \right) \left( \matrix{ m - k \cr m - k \cr} \right)x^{\,k} z^{\,m} } } = \cr & = \sum\limits_{0\, \le \,k\,} {\left( \matrix{ r \cr k \cr} \right)\left( {zx} \right)^{\,\,k} \sum\limits_{0\, \le \,\,m - k} {\left( \matrix{ m - k \cr m - k \cr} \right)z^{\,m - k} } } = \cr & = {1 \over {1 - z}}\sum\limits_{0\, \le \,k\,} {\left( \matrix{ r \cr k \cr} \right)\left( {zx} \right)^{\,k} } = {{\left( {1 + zx} \right)^{\,r} } \over {\left( {1 - z} \right)}} \cr} } \tag {a.1}$$

Since the sum in $x^k$ is limited, it will converge for any value of $x$.
Concerning the sum in $z$ we consider it as a formal sum.
So $$ B_{\,T} (z,r,m) = \left[ {z^{\,m} } \right]F_{\,a} (x,z,r) $$

b) Through the Beta function

The sum of the binomial truncated to $m$ can be given an integral representation which leads to the Incomplete Regularized Beta Function. $$ \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,m} {\left( \matrix{ r \cr k \cr} \right)x^{\,r - k} \left( {1 - x} \right)^{\,k} } = \left( {r - m} \right)\left( \matrix{ r \cr m \cr} \right)\int_0^x {t^{\,r - m - 1} \left( {1 - t} \right)^{\,m} dt} = \cr & = \left( {r - m} \right){{\Gamma (r + 1){\rm B}(x;\,r - m,m + 1)} \over {\Gamma (m + 1)\Gamma (r - m + 1)}} = {{{\rm B}(x;\,r - m,m + 1)} \over {{\rm B}(r - m,m + 1)}} = \cr & = CDF\;Beta\;Distrib.\quad \left| \matrix{ \;r,x \in \mathbb R \hfill \cr \;x \in \left[ {0,1} \right] \hfill \cr \;0 \le m \in \mathbb Z < r \hfill \cr} \right. \cr} $$ re. to this post

Making the substitution $$ {{1 - x} \over x} = y\quad \Rightarrow \;\quad {1 \over {1 + y}} = x\quad \left| {\;x \in \left[ {0,1} \right]\; \leftrightarrow \;y \in \left[ {0,\infty } \right)} \right. $$ we get $$ \bbox[lightyellow] { \eqalign{ & B_{\,T} (y,r,m) = \sum\limits_{0\, \le \,k\, \le \,m} {\left( \matrix{ r \cr k \cr} \right)y^{\,k} } = \cr & = \left( {1 + y} \right)^{\,r} {{{\rm B}\left( {{1 \over {1 + y}};\,r - m,m + 1} \right)} \over {{\rm B}\left( {r - m,m + 1} \right)}} \quad \quad \left| \matrix{ \,y \in \left[ {0,\infty } \right) \hfill \cr \;0 \le m \in \mathbb Z \hfill \cr \,\,0 \le m < r \in \mathbb R \hfill \cr} \right. \cr} } \tag {b.1}$$

Yet, since the validity of the above is limited to non-negative values of $y$, it is not useful to operate the bi-section.

c) Through the Hypergeometric function

Either from the expression of the Inc. Rat. Beta wrt the Hypergeometric function, either directly we obtain $$ \eqalign{ & B_{\,T} (z,r,m) = \sum\limits_{k\, = \,0}^m {\left( \matrix{ r \cr k \cr} \right)z^{\,k} } = z^{\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {\left( \matrix{ r \cr m - k \cr} \right)\left( {{1 \over z}} \right)^{\,k} } = \cr & = z^{\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {{{r^{\,\underline {\,m - k\,} } } \over {1^{\,\overline {\,m - k\;} } }}\left( {{1 \over z}} \right)^{\,k} } = z^{\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {{{r^{\,\underline {\,m\,} } \left( {r - m} \right)^{\,\underline {\, - k\,} } } \over {1^{\,\overline {\,m\;} } \left( {m + 1} \right)^{\,\overline {\, - k\;} } }}\left( {{1 \over z}} \right)^{\,k} } = \cr & = z^{\,m} \left( \matrix{ r \cr m \cr} \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {{{\left( {r - m} \right)^{\,\underline {\, - k\,} } } \over {\left( {m + 1} \right)^{\,\overline {\, - k\;} } }} \left( {{1 \over z}} \right)^{\,k} } = \cr & = z^{\,m} \left( \matrix{ r \cr m \cr} \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {{{m^{\,\underline {\,k\,} } } \over {\left( {r - m + 1} \right)^{\,\overline {\,k\;} } }} \left( {{1 \over z}} \right)^{\,k} } = \cr & = z^{\,m} \left( \matrix{ r \cr m \cr} \right)\sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {{{\left( { - m} \right)^{\,\overline {\,k\;} } } \over {\left( {r - m + 1} \right)^{\,\overline {\,k\;} } }} \left( { - {1 \over z}} \right)^{\,k} } = \cr & = z^{\,m} \left( \matrix{ r \cr m \cr} \right) {}_2F_{\,1} \left( {\left. {\matrix{ { - m,\;1} \cr {r + 1 - m} \cr } \;} \right|\; - {1 \over z}} \right) \cr} $$ where $x^{\,\underline {\,k\,} } ,\quad x^{\,\overline {\,k\,} } $ represent respectively the Falling and Rising Factorial.
Summarizing we have $$ \bbox[lightyellow] { \eqalign{ & B_{\,T} (z,r,m) = \sum\limits_{k\, = \,0}^m {\left( \matrix{ r \cr k \cr} \right)z^{\,k} } = z^{\,m} \sum\limits_{0\, \le \,k\,\left( { \le \,m} \right)} {\left( \matrix{ r \cr m - k \cr} \right)\left( {{1 \over z}} \right)^{\,k} } = \cr & = z^{\,m} \left( \matrix{ r \cr m \cr} \right){}_2F_{\,1} \left( {\left. {\matrix{ { - m,\;1} \cr {r + 1 - m} \cr } \;} \right|\; - {1 \over z}} \right) = \cr & = \left( {1 + z} \right)^{\,m} \left( \matrix{ r \cr m \cr} \right) {}_2F_{\,1} \left( {\left. {\matrix{ { - m,\;r - m} \cr {r + 1 - m} \cr } \;} \right|\;{1 \over {1 + z}}} \right) \quad \left| \matrix{ \;z \in \mathbb C \hfill \cr \;m \in \mathbb Z \hfill \cr \;r \notin \mathbb Z\; \vee \;m \le r \in \mathbb Z \hfill \cr} \right. \cr} } \tag {c.1}$$ where the last equivalent expression comes from the Beta function, as well as from one of the Pfaff transformations. For the bisection we will use the first.

We can also directly express the bi-sected binomial as a ${}_3F_{\,2} $ Hypergeometric function.
Starting from $$ \sum\limits_{0\, \le \,k\, \le \,m} {\left( \matrix{ r \cr 2k + p \cr} \right)y^{\,2k + p} } = y^{\,2m + p} \sum\limits_{0\, \le \,k\left( {\, \le \,m} \right)} {\left( \matrix{ r \cr 2m + p - 2k \cr} \right)\left( {y^{\, - 2} } \right)^{\,k} } $$ Putting $$ t_{\,k} = \left( \matrix{ r \cr 2m + p - 2k \cr} \right) = {{r^{\,\underline {\,2m + p - 2k\,} } } \over {1^{\,\overline {\,2m + p - 2k\,} } }} $$ we have $$ \eqalign{ & t_{\,0} = \left( \matrix{ r \cr 2m + p \cr} \right) \cr & {{t_{\,k + 1} } \over {t_{\,k} }} = {{r^{\,\underline {\,2m + p - 2 - 2k\,} } } \over {1^{\,\overline {\,2m + p - 2 - 2k\,} } }} {{1^{\,\overline {\,2m + p - 2k\,} } } \over {r^{\,\underline {\,2m + p - 2k\,} } }} = \cr & = {{\left( { - r} \right)^{\,\overline {\,2m + p - 2 - 2k\,} } } \over {\left( { - r} \right)^{\,\overline {\,2m + p - 2k\,} } }} {{1^{\,\overline {\,2m + p - 2k\,} } } \over {1^{\,\overline {\,2m + p - 2 - 2k\,} } }} = \cr & = {{\left( {1\, + 2m + p - 2 - 2k\,} \right)^{\,\overline {\,2\,} } } \over {\left( { - r\, + 2m + p - 2 - 2k\,} \right)^{\,\overline {\,2\,} } }} = \cr & = {{\left( {2m + p - 1 - 2k\,} \right)\left( {\,2m + p - 2k\,} \right)} \over {\left( { - r\, + 2m + p - 2 - 2k\,} \right)\left( { - r\, + 2m + p - 1 - 2k\,} \right)}} = \cr & = {{\left( {k - m - \left( {p - 1} \right)/2\,} \right)\left( {\,k - m - p/2\,} \right)} \over {\left( {k - m + \left( {r - p + 2} \right)/2\,} \right)\left( {k - m + \left( {r - p + 1} \right)/2\,} \right)}} \cr} $$ and so we get $$ \bbox[lightyellow] { \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,m} {\left( \matrix{ r \cr 2k + p \cr} \right)z^{\,2k + p} } = \cr & = z^{\,2m + p} \sum\limits_{0\, \le \,k\left( {\, \le \,m} \right)} {\left( \matrix{ r \cr 2m + p - 2k \cr} \right)\left( {z^{\, - 2} } \right)^{\,k} } \cr & = z^{\,2m + p} \left( \matrix{ r \cr 2m + p \cr} \right) {}_3F_{\,2} \left( {\left. {\matrix{ { - m - \left( {p - 1} \right)/2,\; - m - p/2,\;1} \cr { - m + \left( {r - p + 2} \right)/2,\; - m + \left( {r - p + 1} \right)/2} \cr } \;} \right|\;{1 \over {z^{\,2} }}} \right) \cr & \left| \matrix{ \;z \in \mathbb R \hfill \cr \;0 \le m,\, p \in \mathbb Z \hfill \cr \;2m + p \le r \in \mathbb R \hfill \cr} \right. \cr} } \tag {c.2}$$