A Short Proof of Knuth's Old Sum

1. Introduction

There appears to be a renewed interest [1,2,5,8,10] 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 [7] 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 [9] 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 () follows from setting $n=1$ in (1.2).

In Section 5, we will derive the following complements of Knuth’s old sum:

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

and

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

(1.5)

Identity (1.5) is the famous combinatorial identity concerning the convolution of central binomial coefficients. Many different proofs of this identity exist in the literature, (see Mikić [6] and the many references therein).

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,

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

where v is a real number; 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 7, some presumably new polynomial identities, including the following:

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

(1.6)

Identity (1.6) subsumes Knuth’s old sum (1.1) (at $x=0$), as well as (1.3) (at $x=1$).

Finally, in Section 8, 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

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

(2.1)

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

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

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)

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

(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

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

(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 [4] 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

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

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

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

(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

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

(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:

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

and

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

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

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

(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. Complements of Knuth’s Old Sum

Theorem 7. 

If n is a non-negative integer, then

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

(5.1)

Proof. 

Set $a={cos}^2(x/2)$ and $b=-{sin}^2(x/2)$ in the binomial theorem

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

(5.2)

to obtain

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

(5.3)

from which (5.1) follows by term-wise integration using Lemma 1. □

Theorem 8. 

If n is a non-negative integer, then

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

Proof. 

Set $a={cos}^2(x/2)$ and $b={sin}^2(x/2)$ in the binomial theorem (29) to obtain

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

(5.4)

from which the stated identity follows by term-wise integration using Lemma 1. □

Next, we present a generalization of (5.1).

Theorem 9. 

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

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

(5.5)

Proof. 

Multiply through (5.3) by ${sin}^{v}x$ and integrate from 0 to $\pi$, using Lemma 1. □

We conclude this section with a generalization of (5).

Theorem 10. 

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

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

(5.6)

Proof. 

Multiply through (5.4) by ${sin}^{v}x$ and integrate from 0 to $\pi$, using Lemma 1. □

6. Combinatorial Identities Associated with Polynomial Identities of a Certain Type

In this section we derive the combinatorial identities associated with any polynomial identity having the following form:

$$P(t,\dots )=\sum _{k=s}^{n}f\left(k\right){\left(1+t\right)}^{p\left(k\right)}=\sum _{k=m}^{r}g\left(k\right)t^{q\left(k\right)};$$

(6.1)

where m, n, r and s are non-negative integers, $p\left(k\right)$ and $q\left(k\right)$ are sequences of non-negative integers, $f\left(k\right)$ and $g\left(k\right)$ are sequences, and t is a complex variable.

Theorem 11. 

Let $P(t,\dots )$ be the polynomial identity given in (6.1). Let v be an arbitrary real number.

• Suppose that, for every integer j, each of $q\left(2j\right)$ and $q(2j-1)$ is a sequence of positive integers having a definite parity but such that the parity of $q\left(2j\right)$ is different from the parity of $q(2j-1)$ for every integer j.
  If $q\left(2j\right)$ is an even integer for every integer j, then

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

  (6.2)

  while if $q\left(2j\right)$ is an odd integer for every integer j, then

  $$\begin{array}{cc}\hfill & \sum _{k=s}^n{\displaystyle \frac{f\left(k\right)}{2^{p\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2p\left(k\right)+v}{\left(2p\left(k\right)+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{p\left(k\right)+v}{v/2}}\right)}^{-1}\hfill \\ & =\sum _{k=⌊(m+2)/2⌋}^{⌈r/2⌉}{\displaystyle \frac{g(2k-1)}{2^{q(2k-1)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{q(2k-1)}{q(2k-1)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(q(2k-1)+v\right)/2}{v/2}}\right)}^{-1}.\hfill \end{array}$$

  (6.3)
• Suppose that, for every integer j, each of $p\left(2j\right)$ and $p(2j-1)$ is a sequence of positive integers having a definite parity but such that the parity of $p\left(2j\right)$ is different from the parity of $p(2j-1)$ for every integer j.
  If $p\left(2j\right)$ is an even integer for every integer j, then

  $$\begin{array}{cc}\hfill & \sum _{k=m}^r{\displaystyle \frac{g\left(k\right){(-1)}^{q\left(k\right)}}{2^{q\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2q\left(k\right)+v}{\left(2q\left(k\right)+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{q\left(k\right)+v}{v/2}}\right)}^{-1}\hfill \\ & =\sum _{k=⌊(s+1)/2⌋}^{⌊n/2⌋}{\displaystyle \frac{{(-1)}^{p\left(2k\right)}f\left(2k\right)}{2^{p\left(2k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{p\left(2k\right)}{p\left(2k\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(p\left(2k\right)+v\right)/2}{v/2}}\right)}^{-1}\hfill \end{array}$$

  (6.4)
  while if $p\left(2j\right)$ is an odd integer for every integer j, then

  $$\begin{array}{cc}\hfill & \sum _{k=m}^r{\displaystyle \frac{g\left(k\right){(-1)}^{q\left(k\right)}}{2^{q\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2q\left(k\right)+v}{\left(2q\left(k\right)+v\right)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{q\left(k\right)+v}{v/2}}\right)}^{-1}\hfill \\ & =\sum _{k=⌊(s+2)/2⌋}^{⌈n/2⌉}{\displaystyle \frac{{(-1)}^{p(2k-1)}f(2k-1)}{2^{p(2k-1)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{p(2k-1)}{p(2k-1)/2}}\right){\left({\displaystyle \genfrac{}{}{0pt}{}{\left(p(2k-1)+v\right)/2}{v/2}}\right)}^{-1}.\hfill \end{array}$$

  (6.5)

Proof. 

Set $t=cosx$ in (6.1) and multiply through by ${sin}^{v}x$ to obtain

$$\begin{array}{cc}\hfill & \sum _{k=s}^n{2}^{p\left(k\right)+v}f\left(k\right){cos}^{2p\left(k\right)+v}\left({\displaystyle \frac{x}{2}}\right){sin}^v\left({\displaystyle \frac{x}{2}}\right)\hfill \\ \hfill & =\sum _{k=⌊(m+1)/2⌋}^{⌊r/2⌋}g\left(2k\right){cos}^{q\left(2k\right)}x{sin}^{v}x+\sum _{k=⌊(m+2)/2⌋}^{⌈r/2⌉}g(2k-1){cos}^{q(2k-1)}x{sin}^{v}x,\hfill \end{array}$$

from which (6.2) and (6.3) follow after term-wise integration from 0 to $\pi$, using Lemma 1. Identities (6.4) and (6.5) are obtained from (6.2) and (6.3) since (6.1) can be written in the following equivalent form:

$$\sum _{k=m}^r{(-1)}^{q\left(k\right)}g\left(k\right){\left(1+t\right)}^{q\left(k\right)}=\sum _{k=s}^n{(-1)}^{p\left(k\right)}f\left(k\right)t^{p\left(k\right)}.$$

□

In particular,

• Suppose that, for every integer j, each of $q\left(2j\right)$ and $q(2j-1)$ is a sequence of positive integers having a definite parity but such that the parity of $q\left(2j\right)$ is different from the parity of $q(2j-1)$ for every integer j.
  If $q\left(2j\right)$ is an even integer for every integer j, then

  $$\sum _{k=s}^n{\displaystyle \frac{f\left(k\right)}{2^{p\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2p\left(k\right)}{p\left(k\right)}}\right)=\sum _{k=⌊(m+1)/2⌋}^{⌊r/2⌋}{\displaystyle \frac{g\left(2k\right)}{2^{q\left(2k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{q\left(2k\right)}{q\left(2k\right)/2}}\right),$$

  (6.6)

  while if $q\left(2j\right)$ is an odd integer for every integer j, then

  $$\sum _{k=s}^n{\displaystyle \frac{f\left(k\right)}{2^{p\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2p\left(k\right)}{p\left(k\right)}}\right)=\sum _{k=⌊(m+2)/2⌋}^{⌈r/2⌉}{\displaystyle \frac{g(2k-1)}{2^{q(2k-1)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{q(2k-1)}{q(2k-1)/2}}\right),$$

  (6.7)
• Suppose that, for every integer j, each of $p\left(2j\right)$ and $p(2j-1)$ is a sequence of positive integers having a definite parity but such that the parity of $p\left(2j\right)$ is different from the parity of $p(2j-1)$ for every integer j.
  If $p\left(2j\right)$ is an even integer for every integer j, then

  $$\sum _{k=m}^r{\displaystyle \frac{g\left(k\right){(-1)}^{q\left(k\right)}}{2^{q\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2q\left(k\right)}{q\left(k\right)}}\right)=\sum _{k=⌊(s+1)/2⌋}^{⌊n/2⌋}{\displaystyle \frac{{(-1)}^{p\left(2k\right)}f\left(2k\right)}{2^{p\left(2k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{p\left(2k\right)}{p\left(2k\right)/2}}\right),$$

  (6.8)

  while if $p\left(2j\right)$ is an odd integer for every integer j, then

  $$\sum _{k=m}^r{\displaystyle \frac{g\left(k\right){(-1)}^{q\left(k\right)}}{2^{q\left(k\right)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2q\left(k\right)}{q\left(k\right)}}\right)=\sum _{k=⌊(s+2)/2⌋}^{⌈n/2⌉}{\displaystyle \frac{{(-1)}^{p(2k-1)}f(2k-1)}{2^{p(2k-1)}}}\left({\displaystyle \genfrac{}{}{0pt}{}{p(2k-1)}{p(2k-1)/2}}\right).$$

  (6.9)

Corollary 12. 

Let an arbitrary polynomial identity have the following form:

$$P(t,\dots )=\sum _{k=s}^{n}f\left(k\right){\left(1+t\right)}^k=\sum _{k=m}^{r}g\left(k\right)t^k,$$

(6.10)

where m, n, r and s are non-negative integers, $p\left(k\right)$ and $q\left(k\right)$ are sequences of non-negative integers, $f\left(k\right)$ and $g\left(k\right)$ are sequences, and t is a complex variable. Let v be an arbitrary real number. Then

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

(6.11)

and

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

(6.12)

In particular,

$$\sum _{k=s}^n{\displaystyle \frac{f\left(k\right)}{2^k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\sum _{k=⌊(m+1)/2⌋}^{⌊r/2⌋}{\displaystyle \frac{g\left(2k\right)}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right),$$

(6.13)

and

$$\sum _{k=m}^r{\displaystyle \frac{g\left(k\right){(-1)}^k}{2^k}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right)=\sum _{k=⌊(s+1)/2⌋}^{⌊n/2⌋}{\displaystyle \frac{f\left(2k\right)}{2^{2k}}}\left({\displaystyle \genfrac{}{}{0pt}{}{2k}{k}}\right).$$

(6.14)

7. Polynomial Identities

In this section, by following the procedure outlined in Section 6, we derive new polynomial identities associated with the binomial theorem.

Theorem 13. 

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

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

(7.1)

Proof. 

Consider the following variation on the binomial theorem:

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

This identity has the form of (6.10). Use (6.11) with

$$f\left(k\right)={(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){(1-x)}^{n-k},g\left(k\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)x^{n-k},s=0=m,r=n,$$

to obtain (7.1). □

Corollary 14. 

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

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

(7.2)

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

(7.3)

Proof. 

Identity (1.6) on page 2 and identities (7.2) and (7.3) correspond to the evaluation of (7.1) at $v=0$, $v=1$ and $v=2$, respectively.

In deriving (7.2), 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 15. 

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

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

(7.4)

Proof. 

Consider another variation on the binomial theorem:

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

Again, this identity has the form of (6.10). Use (6.11) with

$$f\left(k\right)={\displaystyle \frac{n}{k}}{(1-x)}^k{x}^{n-k},g\left(k\right)={\displaystyle \frac{n}{k}}{(1-x)}^k,s=0=m,r=n,$$

to obtain (7.4). □

Corollary 16. 

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

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

(7.5)

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

(7.6)

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

(7.7)

Proof. 

Identities (7.5), (7.6) and (7.7) correspond to the evaluation of (7.4) at $v=0$, $v=1$ and $v=2$, respectively. □

8. More Combinatorial Identities

8.1. Identities from the Binomial Theorem

Theorem 17. 

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

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

(8.1)

and

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

(8.2)

Proof. 

Evaluate (7.1) at $x=-1$ and $x=2$, respectively. □

Remark 2. 

Setting $x=0$ in (7.1) reproduces identity (4.2) while setting $x=1$ reproduces (4.3).

Proposition 18. 

If n is a non-negative integer, then

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

(8.3)

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

(8.4)

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

(8.5)

Proof. 

Set $x=-1$ in each of identities (1.6)– (7.3). □

Proposition 19. 

If n is a non-negative integer, then

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

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

(8.6)

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

(8.7)

Proof. 

Set $x=1$ in each of identities (1.6)– (7.3). □

Proposition 20. 

If n is a non-negative integer, then

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

(8.8)

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

(8.9)

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

(8.10)

Proof. 

Set $x=2$ in each of identities (1.6)–(7.3). □

Theorem 11. 

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

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

(8.11)

Proof. 

Set $x=-1$ in (7.4). □

Proposition 22. 

If n is a non-negative integer, then

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

(8.12)

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

(8.13)

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

(8.14)

Proof. 

Set $x=-1$ in each of identities (7.5)– (7.7) or what is the same thing, $v=0$, $v=1$ and $v=2$ in (8.11). □

8.2. Identities from Waring’s Formulas

Waring’s formula and its dual [3] Equations (22) and (1) are

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

(8.15)

and

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

(8.16)

Identity (8.15) holds for positive integer n while identity (8.16) holds for any non-negative integer n.

Theorem 12. 

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

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

(8.17)

In particular,

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

(8.18)

Proof. 

Write ${cos}^2(x/2)$ for x and ${sin}^2(y/2)$ for y in (8.15) and multiply through by ${sin}^{v}x$ to obtain

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

from which upon term-wise integration from 0 to $\pi$, identity (8.17) follows. □