A Short Proof of Knuth's Old Sum

1. Introduction

There appears to be a renewed interest [1,2,4,6,8] in the famous Knuth’s old sum (also known as the Reed Dawson identity),

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\left\{\begin{array}{cc}{2}^{-n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n/2}}\right),\hfill & \mathrm{if}n\mathrm{is}\mathrm{even};\hfill \\ 0,\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.$$

(1.1)

Many different proofs of this identity and various generalizations exist in the literature (see [5] for a survey).

In this paper we give a very short proof of (1.1) and offer the following generalization:

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k-m}\left({\displaystyle \genfrac{}{}{0pt}{}{2(k+m)}{k+m}}\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}\sum _{k=0}^{⌊m/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k}}\right)2^{-n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+n}{(2k+n)/2}}\right),\hfill & \mathrm{if}n\mathrm{is}\mathrm{even};\hfill \\ -\sum _{k=1}^{⌈m/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k-1}}\right)2^{-n-2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+n-1}{(2k+n-1)/2}}\right),\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd};\hfill \end{array}\right.\hfill \end{array}$$

(1.2)

where m and n are non-negative integers and, as usual, $\lfloor z\rfloor$ is the greatest integer less than or equal to z while $\lceil z\rceil$ is the smallest integer greater than or equal to z.

The following special cases of (1.2) were also reported in Riordan [7], p.72, Problem 4(b)]:

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right),\end{array}$$

(1.3)

$$\begin{array}{c}\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right)2^{n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)={\displaystyle \frac{1}{2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n+2}{n+1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)={\displaystyle \frac{n}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right).\end{array}$$

(1.4)

Identity (1.3) corresponds to setting $n=0$ in (1.2) and re-labeling m as n; while (1.4) follows from setting $n=1$ in (1.2).

Identity (1.2) is itself a particular case of a more general identity, stated in Theorem 2, which has many interesting consequences, including another generalization of Knuth’s old sum, namely,

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}=\left\{\begin{array}{cc}{2}^{-n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(n+v\right)/2}{v/2}}\right)}^{-1},\hfill & \mathrm{if}n\mathrm{is}\mathrm{even};\hfill \\ 0,\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd};\hfill \end{array}\right.$$

for a real number v, as well as simple, apparently new combinatorial identities such as

$$\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right)2^{n-2k}{C}_k={\displaystyle \frac{1}{2}}{C}_{n+2}-C_{n+1};$$

where, here and throughout this paper,

$$C_j={\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{2j}{j}\right)}{j+1}},$$

defined for every non-negative integer j, is a Catalan number.

Based on the binomial theorem, we will derive, in Section 5, some presumably new polynomial identities, including the following:

$$\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left(1-x\right)}^{n-k}=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)x^{n-2k}.$$

(1.5)

Identity (1.5) subsumes Knuth’s old sum (1.1) (at $x=0$), as well as (1.3) (at $x=1$).

Finally, in Section 6, the polynomial identities will facilitate the derivation of apparently new combinatorial identities such as

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\displaystyle \frac{1}{2k+1}}={\displaystyle \frac{2^{n-1}}{2^n-1}}\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right){\displaystyle \frac{1}{k}},n\ne 0,\\ \sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}{C}_k={\displaystyle \frac{2^{-n+1}}{n+2}}\left(2n+1\right)C_n,\end{array}$$

and

$$\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2\left(2k+1\right)}{k+2}}{C}_k=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)C_k.$$

2. Required Identities

In order to give the short proof of Knuth’s old sum, we need a couple of definite integrals which we establish in Lemma 1.

The binomial coefficients are defined, for non-negative integers m and n, by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n}}\right)=\left\{\begin{array}{cc}{\displaystyle \frac{m!}{n!(m-n)!}},\hfill & m\ge n;\hfill \\ 0,\hfill & m<n;\hfill \end{array}\right.$$

the number of distinct sets of n objects that can be chosen from m distinct objects.

Generalized binomial coefficients are defined for complex numbers u and v, excluding the set of negative integers, by

$$\left({\displaystyle \genfrac{}{}{0pt}{}{u}{v}}\right)={\displaystyle \frac{\Gamma \left(u+1\right)}{\Gamma \left(v+1\right)\Gamma \left(u-v+1\right)}},$$

(2.1)

where $\Gamma \left(z\right)$ is the Gamma function defined by

$$\Gamma \left(z\right)={\int }_0^{\infty }{e}^{-t}{t}^{z-1}dt={\int }_0^{\infty }{\left(log\left(1/t\right)\right)}^{z-1}dt$$

and extended to the rest of the complex plain, excluding the non-positive integers, by analytic continuation.

Lemma 1.

Let u and v be complex numbers such that $\Re u>-1$ and $\Re v>-1$. Let m be a non-negative integer. Then

$${\int }_0^{\pi }{cos}^u(x/2)dx=2^{-u}\pi \left({\displaystyle \genfrac{}{}{0pt}{}{u}{u/2}}\right)={\int }_0^{\pi }{sin}^u(x/2)dx,$$

(2.2)

$${\int }_0^{\pi }{cos}^{m}xdx=\left\{\begin{array}{cc}{2}^{-m}\pi \left({\displaystyle \genfrac{}{}{0pt}{}{m}{m/2}}\right),\hfill & \mathit{if}m\mathit{is}\mathit{even};\hfill \\ 0,\hfill & \mathit{if}m\mathit{is}\mathit{odd};\hfill \end{array}\right.$$

(2.3)

and, more generally,

$$I(u,v):={\int }_0^{\pi }{cos}^u\left({\displaystyle \frac{x}{2}}\right){sin}^v\left({\displaystyle \frac{x}{2}}\right)dx=2^{-u-v}\pi \left({\displaystyle \genfrac{}{}{0pt}{}{u}{u/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{v}{v/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(u+v\right)/2}{u/2}}\right)}^{-1},$$

(2.4)

and

$$J(m,v):={\int }_0^{\pi }{cos}^{m}x{sin}^{v}xdx=\left\{\begin{array}{cc}{2}^{-m-v}\pi \left({\displaystyle \genfrac{}{}{0pt}{}{m}{m/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{v}{v/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(m+v\right)/2}{m/2}}\right)}^{-1},\hfill & \mathit{if}m\mathit{is}\mathit{even};\hfill \\ 0,\hfill & \mathit{if}m\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

(2.5)

Obviously $I(v,u)=I(u,v)$, a symmetry property that is not possessed by $J(m,v)$.

Proof. 

Identities (2.4) and (2.5) are immediate consequences of the well-known Beta function integral ([3] Entry 3.621.5, Page 397):

$$K(u,v):={\int }_0^{\pi /2}{cos}^{u}x{sin}^{v}xdx=2^{-u-v-1}\pi \left({\displaystyle \genfrac{}{}{0pt}{}{u}{u/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{v}{v/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(u+v\right)/2}{u/2}}\right)}^{-1},$$

(2.6)

valid for $\Re u>-1$, $\Re v>-1$, with the symmetry property $K(u,v)=K(v,u)$.

Identity (2.4) is obtained via a simple change of the integration variable from x to y in (2.6), with $x=y/2$.

To prove (2.5), write

$$J(m,v)={\int }_0^{\pi }{cos}^{m}x{sin}^{v}xdx={\int }_0^{\pi /2}{cos}^{m}x{sin}^{v}xdx+{\int }_{\pi /2}^{\pi }{cos}^{m}x{sin}^{v}xdx.$$

Change the integration variable in the second integral on the right hand side from x to y via $x=y+\pi /2$; this gives

$$\begin{array}{cc}\hfill J(m,v)& ={\int }_0^{\pi /2}{cos}^{m}x{sin}^{v}xdx+{(-1)}^m{\int }_0^{\pi /2}{sin}^{m}y{cos}^{v}ydy\hfill \\ \hfill & =K(m,v)+{(-1)}^{m}K(v,m)\hfill \\ \hfill & =\left(1+{(-1)}^m\right)K(m,v);\hfill \end{array}$$

and hence (2.5).

□

Remark 1.

Since, for a real number u,

$$1+{(-1)}^u=2{cos}^2\left({\displaystyle \frac{\pi u}{2}}\right)+isin\left(\pi u\right),$$

the $J(m,v)$ stated in (2.5) is a special case of the following more general result:

$$\begin{array}{cc}\hfill J\left(u,v\right)& ={\int }_0^{\pi }{cos}^{u}x{sin}^{v}xdx\hfill \\ \hfill & ={\displaystyle \frac{\pi }{2^{u+v+1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{u}{u/2}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{v}{v/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(u+v\right)/2}{u/2}}\right)}^{-1}\left(2{cos}^2\left({\displaystyle \frac{\pi u}{2}}\right)+isin\left(\pi u\right)\right),\hfill \end{array}$$

(2.7)

which is valid for $u>-1$ and $\Re v>-1$.

3. A Short Proof of Knuth’s Old Sum

Theorem 1.

If n is a non-negative integer, then

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\left\{\begin{array}{cc}{2}^{-n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n/2}}\right),\hfill & \mathit{if}n\mathit{is}\mathit{even};\hfill \\ 0,\hfill & \mathit{if}n\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

Proof. 

Substitute $-cosx-1$ for y in the binomial theorem

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)y^k={(1+y)}^n,$$

to obtain

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^k{cos}^{2k}(x/2)={(-1)}^n{cos}^{n}x.$$

(3.1)

Thus

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^k{\int }_0^{\pi }{cos}^{2k}(x/2)dx={(-1)}^n{\int }_0^{\pi }{cos}^{n}xdx,$$

and hence (1.1) on account of (2.2) and (2.3). □

4. A Generalization of Knuth’s Old Sum

In this section we extend (1.1) by introducing an arbitrary non-negative integer m and a real number v.

Theorem 2.

If m and n are non-negative integers and v is a real number, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k-m}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+2m+v}{(2k+2m+v)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+m+v}{v/2}}\right)}^{-1}\hfill \\ \hfill & =\left\{\begin{array}{cc}\sum _{k=0}^{⌊m/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k}}\right)2^{-n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+n}{(2k+n)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{(2k+n+v)/2}{(2k+n)/2}}\right)}^{-1},\hfill & \mathit{if}n\mathit{is}\mathit{even};\hfill \\ -\sum _{k=1}^{⌈m/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k-1}}\right)2^{-n-2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+n-1}{(2k+n-1)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{(2k+n-1+v)/2}{(2k+n-1)/2}}\right)}^{-1},\hfill & \mathit{if}n\mathit{is}\mathit{odd}.\hfill \end{array}\right.\hfill \end{array}$$

(4.1)

In particular,

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k-m}\left({\displaystyle \genfrac{}{}{0pt}{}{2(k+m)}{k+m}}\right)\hfill \\ \hfill & =\left\{\begin{array}{cc}\sum _{k=0}^{⌊m/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k}}\right)2^{-n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+n}{(2k+n)/2}}\right),\hfill & \mathrm{if}n\mathrm{is}\mathrm{even};\hfill \\ -\sum _{k=1}^{⌈m/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k-1}}\right)2^{-n-2k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+n-1}{(2k+n-1)/2}}\right),\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.\hfill \end{array}$$

Proof. 

Since

$${\left(1+cosx\right)}^m=2^m{cos}^{2m}\left({\displaystyle \frac{x}{2}}\right)=\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){cos}^{k}x$$

and

$${sin}^{v}x=2^v{sin}^v\left({\displaystyle \frac{x}{2}}\right){cos}^v\left({\displaystyle \frac{x}{2}}\right),$$

multiplication of the left hand side of (3.1) by

$$2^{m+v}{cos}^{2m+v}\left({\displaystyle \frac{x}{2}}\right){sin}^v\left({\displaystyle \frac{x}{2}}\right)$$

and the right hand side by

$${sin}^{v}x\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){cos}^{k}x$$

gives

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{k+m+v}{cos}^{2k+2m+v}(x/2){sin}^v(x/2)\hfill \\ \hfill & ={(-1)}^n\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right){cos}^{k+n}x{sin}^{v}x;\hfill \end{array}$$

so that

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{k+m+v}{cos}^{2k+2m+v}(x/2){sin}^v(x/2)\hfill \\ \hfill & ={(-1)}^n\sum _{k=0}^{⌊m/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k}}\right){cos}^{2k+n}x{sin}^{v}x+{(-1)}^n\sum _{k=1}^{⌈m/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k-1}}\right){cos}^{2k-1+n}x{sin}^{v}x,\hfill \end{array}$$

from which (4.1) now follows by termwise integration from 0 to $\pi$, according to the parity of n, using Lemma 1.

□

Corollary 3.

If n is a non-negative integer and v is a real number, then

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}=\left\{\begin{array}{cc}{2}^{-n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{n/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(n+v\right)/2}{v/2}}\right)}^{-1},\hfill & \mathit{if}n\mathit{is}\mathit{even};\hfill \\ 0,\hfill & \mathit{if}n\mathit{is}\mathit{odd};\hfill \end{array}\right.$$

(4.2)

Corollary 4.

If n is a non-negative integer and v is a real number, then

$$\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1}=2^{-n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n+v}{\left(2n+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n+v}{v/2}}\right)}^{-1},$$

(4.3)

and

$$\begin{array}{cc}\hfill & \sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right)2^{n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n+v+2}{\left(2n+v+2\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n+v+1}{v/2}}\right)}^{-1}-\left({\displaystyle \genfrac{}{}{0pt}{}{2n+v}{\left(2n+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{n+v}{v/2}}\right)}^{-1}.\hfill \end{array}$$

(4.4)

Proof. 

Identity (4.3) is obtained by setting $n=0$ in (4.1) and re-labeling m as n while (4.4) is the evaluation of (4.1) at $n=1$ with a re-labeling of m as n. □

Proposition 5.

If n is a non-negative integer, then

$$\begin{array}{c}\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right){\displaystyle \frac{1}{2k+1}}={\displaystyle \frac{2^{n+1}}{n+2}}-{\displaystyle \frac{2^n}{n+1}},\end{array}$$

(4.5)

$$\begin{array}{c}\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right)2^{n-2k}{C}_k={\displaystyle \frac{1}{2}}{C}_{n+2}-C_{n+1}.\end{array}$$

(4.6)

Proof. 

Evaluation of (4.4) at $v=1$ gives (4.5) while evaluation at $v=2$ yields (4.6). In deriving (4.5), we used the following relationships between binomial coefficients:

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{1/2}}\right)={\displaystyle \frac{2^{2r+1}}{\pi }}{\left({\displaystyle \genfrac{}{}{0pt}{}{2r}{r}}\right)}^{-1},\end{array}$$

(4.7)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{r}{r/2}}\right)={\displaystyle \frac{2^{2r}}{\pi }}{\left({\displaystyle \genfrac{}{}{0pt}{}{r}{\left(r-1\right)/2}}\right)}^{-1},\end{array}$$

(4.8)

$$\begin{array}{c}\left({\displaystyle \genfrac{}{}{0pt}{}{r+1/2}{r}}\right)=\left(2r+1\right)2^{-2r}\left({\displaystyle \genfrac{}{}{0pt}{}{2r}{r}}\right),\end{array}$$

(4.9)

and

$$r\left({\displaystyle \genfrac{}{}{0pt}{}{s}{r}}\right)=s\left({\displaystyle \genfrac{}{}{0pt}{}{s-1}{r-1}}\right);$$

(4.10)

all of which can be derived by using the Gamma function identities:

$$\Gamma \left(u+{\displaystyle \frac{1}{2}}\right)=\sqrt{\pi }{2}^{-2u}\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)\Gamma \left(u+1\right),$$

and

$$\Gamma \left(-u+{\displaystyle \frac{1}{2}}\right)={(-1)}^u{2}^{2u}{\left({\displaystyle \genfrac{}{}{0pt}{}{2u}{u}}\right)}^{-1}{\displaystyle \frac{\sqrt{\pi }}{\Gamma \left(u+1\right)}},$$

together with the definition of the generalized binomial coefficients as given in (2.1). □

Proposition 6.

If m and n are non-negative integers, then

$$\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2^{k+m}}{k+m+1}}=\left\{\begin{array}{cc}\sum _{k=0}^{⌊m/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k}}\right){\displaystyle \frac{1}{2k+n+1}},\hfill & \mathit{if}n\mathit{is}\mathit{even};\hfill \\ -\sum _{k=1}^{⌈m/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{2k-1}}\right){\displaystyle \frac{1}{2k+n}},\hfill & \mathit{if}n\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

(4.11)

In particular,

$$\sum _{k=0}^n{\displaystyle \frac{{(-1)}^k\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^k}{k+1}}=\left\{\begin{array}{cc}{\displaystyle \frac{1}{n+1}},\hfill & \mathrm{if}n\mathrm{is}\mathrm{even};\hfill \\ 0,\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd};\hfill \end{array}\right.$$

(4.12)

and

$$\sum _{k=0}^n{\displaystyle \frac{{(-1)}^k\left(\genfrac{}{}{0pt}{}{n}{k}\right)2^{k+1}}{k+2}}=\left\{\begin{array}{cc}{\displaystyle \frac{1}{n+1}},\hfill & \mathrm{if}n\mathrm{is}\mathrm{even};\hfill \\ -{\displaystyle \frac{1}{n+2}},\hfill & \mathrm{if}n\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.$$

(4.13)

Proof. 

Evaluate (4.1) at $v=1$.

□

5. Polynomial Identities

In this section, by employing Lemma 1 we derive new polynomial identities from the binomial theorem.

Theorem 7.

If n is a non-negative integer, v is a real number and x is a complex variable, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}{\left(1-x\right)}^{n-k}\hfill \\ \hfill & =\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1}{x}^{n-2k}.\hfill \end{array}$$

(5.1)

Proof. 

Consider the following variation on the binomial theorem:

$$\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(1+y\right)}^k{\left(1-x\right)}^{n-k}=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)y^k{x}^{n-k}.$$

Write $cosy$ for y, multiply through by ${sin}^{v}y$ and integrate from 0 to $\pi$ with respect to y, using Lemma 1. □

Corollary 8.

If n is a non-negative integer and x is a complex variable, then

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2^k}{k+1}}{\left(1-x\right)}^{n-k}=\sum _{k=0}^{⌊n/2⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{2k}\right)}{2k+1}}{x}^{n-2k},\end{array}$$

(5.2)

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^{n-k}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}{k+2}}{2}^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\left(k+1\right)}{k+1}}\right){\left(1-x\right)}^{n-k}=\sum _{k=0}^{⌊n/2⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{2k}\right)}{k+1}}{2}^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)x^{n-2k}.\end{array}$$

(5.3)

Proof. 

Identity (1.5) on page 2 and identities (5.2) and (5.3) correspond to the evaluation of (5.1) at $v=0$, $v=1$ and $v=2$, respectively.

In deriving (5.2), we used (4.7)–(4.10) to obtain

$$\left({\displaystyle \genfrac{}{}{0pt}{}{2k+1}{k+1/2}}\right)={\displaystyle \frac{2^{4k+4}}{\pi \left(k+1\right)}}{\left({\displaystyle \genfrac{}{}{0pt}{}{2\left(k+1\right)}{k+1}}\right)}^{-2}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+1}{k}}\right),$$

$$\left({\displaystyle \genfrac{}{}{0pt}{}{k+1}{1/2}}\right)={\displaystyle \frac{2^{2k+3}}{\pi }}{\left({\displaystyle \genfrac{}{}{0pt}{}{2\left(k+1\right)}{k+1}}\right)}^{-1},$$

and

$$\left({\displaystyle \genfrac{}{}{0pt}{}{2k+1}{k}}\right)={\displaystyle \frac{1}{2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2\left(k+1\right)}{k+1}}\right).$$

□

Theorem 9.

if n is a non-negative integer, v is a real number and x is a complex variable, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}{\left(1-x\right)}^k{x}^{n-k}\hfill \\ \hfill & =\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1}{(1-x)}^{2k}.\hfill \end{array}$$

(5.4)

Proof. 

Write $cosy$ for y in the following variation on the binomial theorem:

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\left(1-x\right)}^k{\left(1+y\right)}^k{x}^{n-k}=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)y^k{\left(1-x\right)}^k,$$

multiply through by ${sin}^{v}y$ and integrate from 0 to $\pi$ with respect to y, using Lemma 1. □

Corollary 10.

If n is a non-negative integer and x is a complex variable, then

$$\begin{array}{c}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left(1-x\right)}^k{x}^{n-k}=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){(1-x)}^{2k},\end{array}$$

(5.5)

$$\begin{array}{c}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2^k}{k+1}}{\left(1-x\right)}^k{x}^{n-k}=\sum _{k=0}^{⌊n/2⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{2k}\right)}{2k+1}}{(1-x)}^{2k},\end{array}$$

(5.6)

$$\begin{array}{c}\sum _{k=0}^n{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}{k+2}}{2}^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2\left(k+1\right)}{k+1}}\right){\left(1-x\right)}^k{x}^{n-k}=\sum _{k=0}^{⌊n/2⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{2k}\right)}{k+1}}{2}^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){(1-x)}^{2k}.\end{array}$$

(5.7)

Proof. 

Identities (5.5), (5.6) and (5.7) correspond to the evaluation of (5.4) at $v=0$, $v=1$ and $v=2$, respectively.

□

6. More Combinatorial Identities

Theorem 11.

If n is an integer and v is a real number, then

$$\begin{array}{cc}\hfill & \sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1},\hfill \end{array}$$

(6.1)

and

$$\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{n-4k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1}.$$

(6.2)

Proof. 

Evaluate (5.1) at $x=-1$ and $x=2$, respectively. □

Remark 2.

Setting $x=0$ in (5.1) reproduces identity (4.2) while setting $x=1$ reproduces (4.3).

Proposition 12.

If n is a non-negative integer, then

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right),\end{array}$$

(6.3)

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\displaystyle \frac{1}{2k+1}}={\displaystyle \frac{2^{n-1}}{2^n-1}}\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right){\displaystyle \frac{1}{k}},n\ne 0,\end{array}$$

(6.4)

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2k+1}{k+2}}{2}^{n-2k+1}{C}_k=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}{C}_k.\end{array}$$

(6.5)

Proof. 

Set $x=-1$ in each of identities (1.5)–(5.3). □

Proposition 13.

If n is a non-negative integer, then

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=2^{-n}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right),\end{array}$$

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{2k}\right)}{2k+1}}={\displaystyle \frac{2^n}{n+1}},\end{array}$$

(6.6)

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}{\displaystyle \frac{\left(\genfrac{}{}{0pt}{}{n}{2k}\right)}{k+1}}{2}^{-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)={\displaystyle \frac{2^{-n+1}}{n+2}}\left(2n+1\right)C_n.\end{array}$$

(6.7)

Proof. 

Set $x=1$ in each of identities (1.5)–(5.3). □

Proposition 14.

If n is a non-negative integer, then

$$\begin{array}{c}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)2^{-k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{n-4k}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right),\end{array}$$

(6.8)

$$\begin{array}{c}\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\displaystyle \frac{2^{n-2k+1}-2^{2k+1}}{2k+1}}=\sum _{k=1}^{⌈n/2⌉}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k-1}}\right){\displaystyle \frac{2^{2k-1}}{k}},\end{array}$$

(6.9)

$$\begin{array}{c}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2k+1}{k+2}}{2}^{-k+1}{C}_k=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)2^{n-4k}{C}_k.\end{array}$$

(6.10)

Proof. 

Set $x=2$ in each of identities (1.5)–(5.3).

□

Theorem 15.

if n is a non-negative integer and v is a real number, then

$$\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k+v}{\left(2k+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{k+v}{v/2}}\right)}^{-1}=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(2k+v\right)/2}{k}}\right)}^{-1}.$$

(6.11)

Proof. 

Set $x=-1$ in (5.4). □

Proposition 16.

If n is a non-negative integer, then

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n-k}{k}}\right),\end{array}$$

(6.12)

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2^{2k}}{k+1}}=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right){\displaystyle \frac{2^{2k}}{2k+1}},\end{array}$$

(6.13)

$$\begin{array}{c}\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{2\left(2k+1\right)}{k+2}}{C}_k=\sum _{k=0}^{⌊n/2⌋}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{2k}}\right)C_k.\end{array}$$

(6.14)

Proof. 

Set $x=-1$ in each of identities (5.5)–(5.7) or what is the same thing, $v=0$, $v=1$ and $v=2$ in (6.11).

□