200 Combinatorial Identities and Theorems Dataset for LLM finetuning

A curated dataset built for the Discrete Logarithm Neural Network architecture

LeetArxiv is Leetcode for Arxiv papers. We curate high-quality mathematical data.

We need your support to build the Discrete Logarithm Neural Network. Please consider becoming a paying subscriber.

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Frontmatter for the offical paper

We made this dataset for our experimental Discrete Logarithm+Finite Field Neural Network architecture. Here’s a quick recap of our progress.

1.0 Section Recap

• Section 1 : Emulating a GPU on a CPU using Congruences, Prime Numbers and Finite Field Theory.

  • We replaced matrix multiplications with big integer multiplications. Thus, transforming AI into a problem in Number Theory, not Matrix Theory.

• Section 2 : Discrete Logarithm Neural Networks stray even further from Interpretability.

  • We replaced RELU with the discrete logarithm, and Softmax with the modulo operator, completely eliminating Linear Algebra. We fully immersed ourselves in AI as a problem in Number Theory.

Introduction

We have been building a neural network based on Number Theory. We curated the 200 Combinatorial Identities and Theorems dataset to further ‘f around and find out’ with our architecture.

This documents serves as a central resource for hosting combinatorial identities and theorems at the intersection of Number Theory, Finite Field Theory and Combinatorics.

We don’t provide proofs. However, we provide descriptions, latex and code for training neural nets.

Our primary sources include Concrete Mathematics by Donald Knuth¹, Combinatorial Identities by John Riordan², and Combinatorial Identities by Henry Gould³.

You are freely permitted to use this resource to finetune LLMs.

Cite as
Kibicho Murage (2025). Combinatorial Identities and Theorems. LeetArxiv. https://leetarxiv.substack.com/p/oeic.

Combinatorial Identities and Theorems is inspired by Neil J. Sloane’s, Online Encyclopedia of Integer Sequences(OEIS)⁴.

Downloading the Dataset

1. Free readers - We suggest you print the current webpage using Ctrl+P and parse as a PDF or HTML. This can be tokenized for your LLM.

   Sample Downloadable dataset for free readers

   Entry 005: Ordinary Prime Polynomial Identity

   \( \sum_{k=0}^{n} \binom{p-k}{n-k} x^k = x^{p+1} (x-1)^{n-p-1}, \quad \text{provided } 0 \leq p \leq n-1. \)

2. Our financial supporters - We have a comprehensive JSON dataset for our paying subsribers. Send a me private message on Substack with your email address to receive the most up-to-date version of the dataset. Available on March 2nd.
   For each identity, the dataset provides:

   1. entryNumber : The reference number for the identity.

   2. description : A plain-text description of the combinatorial identity.

   3. tags : A list of tags to find related combinatorial identities.

   4. latex : A latex string representing the identity.

   5. imageLink : Link to a png image of the identity.

   6. citation : Source of identity.

   7. codeSample : (If available) A Python or C example of the identity. Feel free to make a code request. I’m doing this in my free time :)

Sample JSON dataset for financial supporters

{
  "entryNumber": "0005",
  "description": "Convert combination involving prime variables into a polynomial with prime degrees",
  "tags": ["finite sum", "prime variables", "polynomials"],
  "latex": "\\sum_{k=0}^{n} \\binom{p-k}{n-k} x^k = x^{p+1} (x-1)^{n-p-1}, \\quad \\text{provided } 0 \\leq p \\leq n-1.",
  "imageLink": "https://leetarxiv.substack.com",
  "codeSample": {
    "C": "#include <stdio.h>\n// C implementation here...",
    "Python": "def oo5():\n    # Python implementation here..."
  },
  "citation":"Gould (1972) page 13. Index 1.96"
}

*This resource changes with time. We’re at 218 identities and theorems. Please subscribe to get updates.

Encyclopedia Entries

Section 0

0.000 Kummer’s Theorem

\(v_p \binom{n}{k} = \sum_{i \geq 0} \frac{\lfloor n / p^i \rfloor - \lfloor k / p^i \rfloor - \lfloor (n-k) / p^i \rfloor}{p}\)

• The highest power of a prime number that divides a binomial coefficient is determined by the number of carries when substracting k from n in base p.

  #include <stdio.h>
  
  // Function to compute the highest power of p dividing binomial(n, k)
  int Zero000(int n, int k, int p)
  {
      int highestPower = 0;
      while (n > 0 || k > 0) 
      {
          int nInBaseP = n % p;
          int kInBaseP = k % p;
          if(nInBaseP < kInBaseP)
          {
              // A carry occurs
              highestPower += 1;  
          }
          n /= p;
          k /= p;
      }
      return highestPower;
  }
  
  int main() 
  {
      int n = 10, k = 3, p = 2;
      printf("(binomial(%d, %d)) = %d ^ %d\n", n, k,p, Zero000(n, k, p));
      return 0;
  }

• Source

0.001 Lucas’ Theorem

\(\binom{n}{k} \equiv \prod_{i \geq 0} \binom{n_i}{k_i} \pmod{p}\)

For a prime number, p, the modulo of a binomial coefficient can be found from the product of congruences mod p, using the base-p expansion of n and k.

#include <stdio.h>

int NchooseK(int n, int k)
{
    if (k > n - k) k = n - k; 
    int result = 1;
    for (int i = 0; i < k; i++) 
    {
        result = result * (n - i) / (i + 1);
    }
    
    return result;
}
// Function to find binomial (n, k) mod p
int Zero001(int n, int k, int p)
{
    int product = 1;
    while (n > 0 || k > 0) 
    {
        int nInBaseP = n % p;
        int kInBaseP = k % p;
        int smallModulo = NchooseK(nInBaseP, kInBaseP) % p;
        product *= smallModulo;
        n /= p;
        k /= p;
    }
    return product % p;
}

int main() 
{
    int n = 1000, k = 300, p = 13;
    printf("(binomial(%d, %d)) = %d (mod %d)\n", n, k, Zero001(n, k, p), p);
    return 0;
}

Source

0.002 Catalan Numbers

\(C_n = \frac{1}{n+1} \binom{2n}{n}\)

Alternatively,

\({\displaystyle C_{n}={2n \choose n}-{2n \choose n+1}} \text{for} \ {\displaystyle n\geq 0\,,}\)

#include <stdio.h>

int NchooseK(int n, int k)
{
    if (k > n - k) k = n - k; 
    int result = 1;
    for (int i = 0; i < k; i++) 
    {
        result = result * (n - i) / (i + 1);
    }
    
    return result;
}
// Function to find the nth catalan number
int Zero002(int n)
{
    int catalanNumber = NchooseK(2*n, n);
    return catalanNumber;
}

int main() 
{
    int n = 4;
    printf("Catalan %d = %d\n", n, Zero002(n));
    return 0;
}

Source

0.003 Sterling number of the first kind

\(\binom{x}{n} = \sum_{k=0}^{n} c_k^n x^k\)

0.004 Sterling number of the second kind

\(x^n = \sum_{k=0}^{n} B_k^n \binom{x}{k}\)

0.005 Gamma function

\(\Gamma(n) = (n-1)!\)

0.006 Lagrange’s Four Square Theorem

\(n = x^2 + y^2 + z^2 + w^2, \quad \forall n \in \mathbb{N}\)

Section 1

This is work in progress

1.1 Vandamonde Convolution

\(\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n} \)

1.1.1 Chu-Vandermonde Identity

\(\sum_{r=0}^{k} \binom{m}{r} \binom{n}{k-r} = \binom{m+n}{k}\)

1.2

\(\sum_{k=0}^{n} \binom{x+k}{k} \binom{y+n-k}{n-k} = \binom{x+y+n+1}{n} \)

1.3

\(\sum_{k=r}^{n-s} \binom{k}{r} \binom{n-k}{s} = \binom{n+1}{r+s+1} \)

1.4

\(\sum_{k=0}^{n} \binom{n}{k} \binom{x+y-n}{x-k} = \binom{x+y}{x} \)

1.5

\(\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \sum_{k=0}^{n} (-1)^k \binom{n-x}{k} \binom{y+n-k}{n} \)

1.6

\(\sum_{k=0}^{n} \binom{x}{k} \binom{x}{2n-k} = \sum_{k=0}^{n} \binom{x}{n-k} \binom{x}{n+k} - \frac{1}{2} \left( \binom{2x}{2n} + \binom{x}{n}^2 \right) \)

1.7

\(\sum_{k=0}^{n-1} \binom{x}{k} \binom{x}{2n-1-k} = \sum_{k=0}^{n-1} \binom{x}{n-1-k} \binom{x}{n+k} = \frac{1}{2} \binom{2x}{2n-1} \)

1.8

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{x}{2k} \binom{x}{n-2k} = \frac{1}{2} \binom{2x}{n} + \frac{(-1)^{n/2}}{2} \binom{x}{n/2} \frac{1+(-1)^n}{2} \)

1.9

\(\sum_{k=0}^{n} \binom{x}{2k} \binom{x}{n-2k} = \frac{1}{2} \binom{2x}{2n} + \frac{(-1)^n}{2} \binom{x}{n} \)

1.10

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{x}{2k+1} \binom{x}{n-2k-1} = \frac{1}{2} \binom{2x}{n} - \frac{(-1)^{n/2}}{2} \binom{x}{n/2} \frac{1+(-1)^n}{2} \)

1.11

\(\sum_{k=0}^{n-1} \binom{x}{2k+1} \binom{x}{2n-2k-1} = \frac{1}{2} \binom{2x}{2n} - \frac{(-1)^n}{2} \binom{x}{n} \)

1.12

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{x}{2k} \binom{2n-x}{n-2k} = \frac{1}{2} \binom{2n}{n} + (-1)^n \frac{2^n}{2} \binom{x-1}{n} \)

1.13

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{x}{2k+1} \binom{2n-x}{n-2k-1} = \frac{1}{2} \binom{2n}{n} - (-1)^n \frac{2^n}{2} \binom{x-1}{n} \)

1.14 Erik Sparre Andersen's Identity

\(\sum_{k=0}^{r} \binom{k}{x} \binom{n-k}{-x} = -\frac{x-r}{n} \binom{x}{n-r-1} - \frac{n-r}{n} \binom{x-1}{r} \binom{-x}{n-r} \)

\(\quad (n \geq 1, \quad 0 \leq r \leq n)\)

\(= -\sum_{k=r+1}^{n} \binom{x}{k} \binom{-x}{n-k} \)

1.15

\(\sum_{k=0}^{r} \binom{x}{k} \binom{1-x}{n-k} = \frac{(n-1)(1-x)-r}{n(n-1)} \binom{x-1}{r} \binom{-x}{n-r-1} \)

\(\quad (n \geq 2, \quad 0 < r \leq n-1) \)

1.16

\(\sum_{k=0}^{n} \binom{-\frac{1}{2}}{k} \binom{\frac{1}{2}}{k} = \binom{-\frac{1}{2}}{n} \binom{-\frac{3}{2}}{n} = (2n+1) \binom{-\frac{1}{2}}{n}^2 \)

1.17

\(\sum_{k=0}^{n} \binom{n}{k} \binom{x}{k} z^k = \sum_{k=0}^{n} \binom{n}{k} \binom{x+n-k}{n} (z-1)^k \)

1.18

\(\sum_{k=0}^{n} \binom{n}{k} \binom{z+k}{k} (x-y)^{n-k} y^k = \sum_{k=0}^{n} \binom{n}{k} \binom{z}{k} x^{n-k} y^k \)

1.19

\(\sum_{k=0}^{n} \binom{n}{k} \binom{x}{k} 2^k = \sum_{k=0}^{n} \binom{n}{k} \binom{x+k}{n} \)

1.20

\(\sum_{k=0}^{n} \binom{n}{k} \binom{k+r}{k} = \binom{n+r}{r} \)

1.21

\(\sum_{k=0}^{n} \binom{x}{k} \binom{y+k}{n-k} 4^k = \sum_{k=0}^{n} \binom{2x}{k} \binom{y}{n-k} 2^k \)

\(= \sum_{k=0}^{n} \binom{2x}{k} \binom{y+2x-k}{n-k} \)

1.22

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{x}{k} \binom{x-k}{n-2k} 2^{2n-2k} = \sum_{k=\lfloor \frac{n}{2} \rfloor}^{n} \binom{x}{k} \binom{n-k}{k} 2^{2k} = 2^n \binom{2x}{n} \)

1.23

\(\sum_{k=0}^{n} (-1)^k \binom{n+x}{n-k} \binom{k+x+1}{k} = \begin{cases} 1, & n=0, \\ -1, & n=1, \\ 0, & n \geq 2. \end{cases} \)

1.24

\(\sum_{k=0}^{n} \binom{x}{2k} \binom{x+n-k-1}{n-k} = \binom{x+2n-1}{2n} \)

1.25

\(\sum_{k=0}^{n} \binom{x}{2k+1} \binom{x+n-k-1}{n-k} = \binom{x+2n}{2n+1} \)

1.26

\(\sum_{k=0}^{n} \binom{2x}{2k} \binom{x-k}{n-k} = \frac{x}{x+n} \binom{x+n}{2n} 2^{2n} \frac{2}{(2n)!} \prod_{k=0}^{n-1} (x^2 - k^2) \)

1.27

\(\sum_{k=0}^{n} \binom{2x+1}{2k+1} \binom{x-k}{n-k} = \frac{2x+1}{2n+1} \binom{x+n}{2n} 2^{2n} \frac{2x+1}{(2n+1)!} \prod_{l=0}^{n-1} [(2x+1)^2 - (2k+1)^2] \)

1.28

\(\sum_{k=0}^{n} \binom{n}{k} \binom{x+k}{x} = \binom{2x+2n}{x+n} \)

1.29

\(\sum_{k=0}^{n} \binom{n}{k} \binom{x}{k-r} = \binom{n+x}{n-r} \)

1.30

\(\sum_{k=0}^{n} \binom{n}{k} \binom{x}{k} k = n \binom{x+n-1}{n} \)

1.31

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{y}{n-k} = \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{x}{k} \binom{y-x}{n-2k} \)

1.32

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{x}{n-k} = (-1)^{n} \frac{n}{2} \binom{x}{n/2} \frac{1+(-1)^n}{2} \)

\(= \binom{x}{n} \frac{2^n (x-n)! \sqrt{\pi}}{(x-n/2)! (-n/2-\frac{1}{2})!} \)

1.33

\(\sum_{k=r}^{n-r} (-1)^k \binom{k}{r} \binom{n-k}{r} = (-1)^r \binom{n/2}{r} \frac{1+(-1)^n}{2} \)

1.34

\(\sum_{k=0}^{2n} (-1)^k \binom{x}{k} \binom{x}{2n-k} = (-1)^n \binom{x}{n} \)

1.35

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{x}{2n-k} = \frac{(-1)^n}{2} \left\{ \binom{x}{n} + \binom{x}{n}^2 \right\} \)

1.36

\(\sum_{k=0}^{n} (-1)^k \binom{x+k}{k} \binom{x+n-k}{n-k} = \binom{x+n/2}{n/2} \frac{1+(-1)^n}{2} \)

1.37

\(\sum_{k=0}^{n} (-1)^k \binom{x}{n-k} \binom{x}{n+k} = \frac{1}{2} \left\{ \binom{x}{n} + \binom{x}{n}^2 \right\} \)

1.38

\(\sum_{k=-n}^{n} (-1)^k \binom{2n}{n-k} \binom{2r}{r-k} = \frac{\binom{2n}{n} \binom{2r}{r}}{\binom{n+r}{n}} = \frac{(2n)! (2r)!}{(n+r)! n! r!} \quad \text{(K. V. Szily)} \)

1.39

\(\sum_{k=0}^{\infty} (-1)^k \binom{x}{k} \binom{-x}{k} = \frac{\sin \pi x}{\pi} \int_0^1 u^{x-1} \frac{(1+u)^x}{1-u} \, du \)

1.40

\(\sum_{k=0}^{n} (-1)^k \binom{-x}{k} \binom{-x}{n-k} = \sum_{k=0}^{n-1} \binom{x}{k+1} \binom{n-1}{k} 2^{k+1} \)

1.41

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{-x}{n-2k} = (-1)^n \binom{x}{n} \)

1.42

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{2n-x}{n-k} = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{x}{k} \binom{2n-2x}{n-2k} \)

\(= (-1)^n \sum_{k=0}^{n} (-1)^k \binom{2n-k}{n-k} \binom{2n-x}{k} 2^k. \)

1.43

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{y-2k}{n-k} 2^k = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{x}{k} \binom{y-2x}{n-2k} \)

1.44

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{2x-2k}{n-k} 2^k = (-1)^n \binom{2x}{n} \)

1.45

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{4x-2k}{n-k} 2^k = (-1)^{\frac{n}{2}} \binom{2x}{n/2} \frac{1+(-1)^n}{2} \)

1.46

\(\sum_{k=0}^{n} (-1)^k \binom{2x+1}{k} \binom{2n-2x-1}{n-k} = (-1)^n 2^{2n} \binom{x}{n} \)

1.47

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{x+k}{r} = (-1)^n \binom{x}{r-n} \)

1.48

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{x+k}{r+k} = (-1)^n \binom{x}{n+r} \)

1.49

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{x-k}{r} = \binom{x-n}{r-n},\)

1.50

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{2n-k}{n} = \binom{2n-x}{n},\)

1.51

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{y-2k}{n-k} 3^k = \sum_{k=0}^{\lfloor n/3 \rfloor} \binom{x}{k} \binom{y-3x}{n-3k},\)

1.52

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{3x-2k}{n-k} 3^k = (-1)^n \binom{x}{n/3} \frac{(-1)^{\lfloor n/3 \rfloor} - (-1)^{\lfloor (n-1)/3 \rfloor}}{2},\)

\(= \begin{cases} 0, & 3 \nmid n, \\ \binom{x}{n/3}, & 3 \mid n. \end{cases}\)

1.53

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{2n-k}{n} 2^k \frac{k}{2n-k} = 2^n \binom{-x/2 + n - 1}{n},\)

1.54

\(\sum_{k=0}^{n} (-1)^k \binom{x}{n-k} \binom{n+k}{k} \frac{n-k}{n+k} \cdot \frac{1}{2^{n+k}} = \binom{x/2}{n},\)

1.55 Generalizes a formula proposed by B.C. Wong1.

\(\sum_{k=0}^{n} (-1)^k \binom{n+1}{k} \binom{2n-2k+x}{n} = \binom{n-x+1}{n}, \quad \\\)

1.56 B.C. Wong

\(\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n+1}{k} \binom{2n-2k}{n} = n+1\)

1.57

\(\sum_{k=0}^{2n} (-1)^k \binom{2n}{k} \binom{2n+2x}{k+x} = (-1)^n \binom{2n}{n} \frac{\binom{2n+2x}{x}}{\binom{n+x}{n}}, \)

1.58

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2x-n}{x-k} = (-1)^{n/2} \frac{1 + (-1)^n}{2} \binom{x}{n/2} \frac{\binom{2x}{x}}{\binom{2x}{n}}, \)

1.59

\(\sum_{k=0}^{2n} (-1)^k \binom{2n}{k} \binom{2x-2n}{x-k} = (-1)^n \binom{x}{n} \frac{\binom{2x}{2n}}{\binom{2x}{2n}}, \)

1.60

\(\sum_{k=0}^{n} (-1)^k \binom{2n}{k} \binom{2x-2n}{x-k} = \frac{(-1)^n}{2} \left\{ \binom{x}{n} + \binom{x}{n}^2 \right\} \frac{\binom{2x}{x}}{\binom{2x}{2n}}, \)

1.61

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{2x-2k}{n-k} 2^k = (-1)^{n/2} \binom{x}{n/2} \frac{1 + (-1)^n}{2}, \)

1.62

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} \binom{2x-2k}{n-k} 2^k = \frac{1 + (-1)^n}{2} \binom{x}{n/2}, \)

1.63

\(\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{x}{k} \binom{2x-2k}{n-2k} = \binom{x}{n} 2^n, \)

1.64

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2k}{j} = (-1)^n \binom{n}{j-n} 2^{2n-j}, \)

1.65

\(\sum_{k=0}^{n} \binom{n}{k}^2 x^k = \sum_{k=0}^{n} \binom{n}{k} \binom{2n-k}{n} (x-1)^k, \)

1.66

\(\sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}, \)

1.67

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{k}^2 = \frac{1}{2} \binom{2n}{n} - \frac{1+(-1)^n}{4} \binom{n}{n/2}^2, \)

1.68

\(\sum_{k=0}^{n} \binom{2n}{k}^2 = \frac{1}{2} \binom{4n}{2n} + \frac{1}{2} \binom{2n}{n}^2, \)

1.69

\(\sum_{k=0}^{n} \binom{2n+1}{k}^2 = \frac{1}{2} \binom{4n+2}{2n+1}, \)

1.70

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}^2 = \frac{1}{2} \binom{2n}{n} + \frac{(-1)^n}{2} \binom{n}{n/2} \frac{1+(-1)^n}{2}, \)

1.71

\(\sum_{k=0}^{n} \binom{2n}{2k}^2 = \frac{1}{2} \binom{4n}{2n} + \frac{(-1)^n}{2} \binom{2n}{n}, \)

1.72

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{2n}{2k}^2 = -\frac{1}{4} \binom{4n}{2n} + \frac{(-1)^n}{4} \binom{2n}{n} + \frac{1+(-1)^n}{4} \binom{2n}{n}^2, \)

1.73

\(\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1}^2 = \frac{1}{2} \binom{2n}{n} - \frac{(-1)^n}{2} \binom{n}{n/2} \frac{1+(-1)^n}{2}, \)

1.74

\(\sum_{k=0}^{\frac{n-1}{2}} \binom{2n}{2k+1}^2 = \frac{1}{2} \binom{4n}{2n} - \frac{(-1)^n}{2} \binom{2n}{n}, \)

1.75

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{2n}{2k+1}^2 = -\frac{1}{4} \binom{4n}{2n} - \frac{(-1)^n}{4} \binom{2n}{n} + \frac{1 - (-1)^n}{4} \binom{2n}{n}^2, \)

1.76

\(\sum_{k=0}^{n} \binom{n}{k}^2 (n - 2k)^2 = n \binom{2n-2}{n-1}, \)

1.77

\(\sum_{k=0}^{n} \binom{n}{k}^2 x^r_k = \sum_{k=0}^{r} \binom{n}{k} \binom{2n-k}{n} B_k^r = s_r, \)

1.78

\(s_0 = \binom{2n}{n}, \quad s_1 = \frac{n}{2} \binom{2n}{n} = (2n-1) \binom{2n-2}{n-1}, \)

1.79

\(s_2 = n^2 \binom{2n-2}{n-1}, \quad s_3 = \frac{n^2 (n+1)}{2} \binom{2n-2}{n-1}, \)

1.80

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = (-1)^{n/2} \binom{n}{n/2} \frac{1+(-1)^n}{2}, \)

or

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n-k}{n} 2^k, \)

1.81

\(\sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^2 = (-1)^n \binom{2n}{n}, \)

1.82

\(\sum_{k=0}^{n} (-1)^k \binom{2n}{k}^2 = \frac{(-1)^n}{2} \left\{ \binom{2n}{n} + \binom{2n}{n}^2 \right\}, \)

1.83

\(\sum_{k=0}^{n} \binom{n}{k}^2 2^k = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \binom{3n - 2k}{2n}. \)

1.84

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2k}{k} \left(\frac{z}{4}\right)^k = \sum_{k=0}^{n} \binom{n}{k} \binom{-\frac{1}{2}}{k} z^k. \)

or

\(= -\frac{1}{2^{2n}} \sum_{k=0}^{n} (-1)^k \binom{2n - 2k}{n - k} \binom{2k}{k} (z - 1)^k. \)

1.85

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2k}{k} \frac{1}{2^{2k}} = \binom{2n}{n} \frac{1}{2^{2n}}. \)

1.86

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2k}{k} = (-1)^n \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \binom{2k}{k}. \)

1.87

\(2^{2n} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2k}{k} = \sum_{k=0}^{n} (-1)^k \binom{2n - 2k}{n - k} \binom{2k}{k} 3^k. \)

1.88

\(2^{2n} \sum_{k=0}^{n} \binom{n}{k} \binom{2k}{k} = \sum_{k=0}^{n} \binom{2n - 2k}{n - k} \binom{2k}{k} 5^k. \)

1.89

\(\sum_{k=0}^{n} (-1)^k (k+1) \binom{n}{k} \binom{2k}{k} \frac{2k+1}{2^{2k}} = \binom{2n}{n} \frac{1}{(2n - 1) 2^{2n}}. \)

1.90

\(\sum_{k=0}^{n} \binom{2n - 2k}{n - k} \binom{2k}{k} = 2^{2n}. \)

1.91

\(\sum_{k=0}^{n} (-1)^k \binom{2n - 2k}{n - k} \binom{2k}{k} = \frac{1 + (-1)^n}{2} \binom{n}{n/2} 2^n, \)

1.92

\(\sum_{k=0}^{n} \binom{2n - 2k}{n - k} \binom{2k}{k} \frac{1}{2k - 1} = \begin{cases} 0, & n \geq 1, \\ -1, & n = 0. \end{cases} \)

1.93

\(\sum_{k=0}^{n} \binom{2n - 2k}{n - k} \binom{2k}{k} \frac{1}{(2k - 1)(2n - 2k - 1)} = \begin{cases} 1, & n = 0, \\ -\frac{4n}{2n + 1} \binom{2n}{n}, & n \geq 1. \end{cases} \)

1.94

\(\sum_{k=0}^{n} \binom{2n - 2k}{n - k} \binom{2k}{k} \frac{1}{(2k - 1)(2n - 2k + 1)} = \frac{2}{2n(2n+1)} \binom{2n}{n}, \quad (n \geq 1). \)

or

\(= 2^n \frac{(2x+1)(2x+3) \cdots (2x+2n-1)}{(x+1) \cdots (x+n)}, \quad (n \geq 1). \)

1.96

\(\sum_{k=0}^{n} \binom{2n - 2k}{n - k} \binom{2k}{k} \frac{1}{2k+1} = \frac{2^{4n}}{(2n+1)} \binom{2n}{n} = \frac{2^{2n}}{(n + \frac{1}{2})} \binom{n}{n}. \)

1.97

\(\sum_{k=0}^{n} \binom{4n - 4k}{2n - 2k} \binom{4k}{2k} = 2^{2n - 1} \binom{2n}{n} + 2^{4n - 1}. \)

1.98

\(\sum_{k=0}^{n-1} \binom{4n - 4k - 2}{2n - 2k - 1} \binom{4k+2}{2k+1} = 2^{4n - 1} - 2^{2n - 1} \binom{2n}{n}, \quad (n \geq 1). \)

1.99

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{k} \binom{2k}{k} 2^{n - 2k} = \binom{2n}{n}. \)

1.100

\(\sum_{k=0}^{n} (-1)^k \binom{n+k}{2k} \binom{2k}{k} \frac{x}{x+k} = (-1)^n \binom{x-1}{n} / \binom{x+n}{n}. \)

1.101

\(\sum_{k=r}^{n} (-1)^k \binom{n}{k} \binom{2k}{k} \binom{k}{k-r} 2^{n-k} = \frac{(-1)^r + (-1)^n}{2} \binom{n}{\frac{n-r}{2}}. \)

1.102

\(\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \binom{2n-2k}{n+k} = 2^{n-r} \binom{n}{r}. \)

1.103 Grosswald’s Identity

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{n+r+k}{k} 2^{n-k} = (-1)^{n/2} \frac{1 + (-1)^n}{2} \binom{n+r}{n/2} \binom{n+2r}{r}. \)

1.104

\(\sum_{k=0}^{n-r} (-1)^k \binom{n}{k+r} \binom{n+k+r}{k} 2^{n-r-k} = (-1)^{\frac{n-r}{2}} \binom{n}{\frac{n-r}{2}} \frac{1 + (-1)^{n-r}}{2}. \)

1.105

\(\sum_{k=0}^{n-r} (-1)^k \binom{n-r}{k} \binom{n+k+r}{n} 2^{n-r-k} = (-1)^{\frac{n-r}{2}} \frac{\binom{n}{r} \binom{n+r}{n}}{\binom{n}{n-r}} \frac{1 + (-1)^{n-r}}{2}. \)

1.106

\(\sum_{k=0}^{2n} (-1)^k \binom{2n}{k} \binom{2n+2k}{k} 3^{2n-k} = \binom{2n}{n}. \)

1.107

\(\sum_{k=r}^{\lfloor n/2 \rfloor} \binom{n-r}{k-r} \binom{n-k}{n-k} 2^{n-2k} = \binom{2n - 2r}{n}. \)

1.108

\(\sum_{k=0}^{n} \binom{k+j}{k} \binom{k+j+m+1}{m} = \sum_{k=0}^{m} \binom{k+j}{k} \binom{k+j+n+1}{n}. \)

1.109

\(\sum_{k=0}^{n} \binom{2k}{k} \binom{2n-k}{n} 2^{2n - 2k} = \binom{4n+1}{2n}. \)

1.110

\(\sum_{k=0}^{n} \binom{2k}{k} \binom{2n-k}{n} \frac{k}{(2n-k) 2^k} = (-1)^n 2^{2n} \binom{-\frac{1}{2}}{n}. \)

1.111

\(\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \binom{2n-2k-1}{n-1} = 1, \quad (n \geq 1). \)

1.112

\(\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n+1}{k} \binom{2n-2k-1}{n} = \frac{1}{2} n(n+1), \quad (n \geq 1). \)

1.113

\(\sum_{k=0}^{\lfloor (r-1)n/r \rfloor} (-1)^k \binom{n+1}{k} \binom{rn-rk}{n} = \binom{n+r-1}{n}. \)

1.114

\(\sum_{k=0}^{n} (-1)^k \binom{2n}{n-k} \binom{2n+2k+1}{2k} = (-1)^n (n+1) 2^{2n}. \)

1.115

\(\sum_{k=0}^{n} \binom{4n+1}{2n-2k} \binom{k+n}{n} = 2^{2n} \binom{3n}{n}. \)

1.116

\(\sum_{k=0}^{n} \binom{4n}{2n-2k} \binom{k+n}{n} = \frac{2}{3} 2^{2n} \binom{3n}{n}. \)

1.117

\(\sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} \binom{2n-2k}{n} = 2^n. \)

1.118

\(\sum_{k=0}^{n} \binom{n}{k} \binom{k}{j} x^k = \binom{n}{j} x^j (1+x)^{n-j}. \)

1.119

\(\sum_{k=j}^{n} (-1)^k \binom{n}{k} \binom{k}{j} = \begin{cases} 0, & j \neq n, \\ (-1)^n, & j = n. \end{cases} \)

1.120

\(\sum_{k=j}^{\lfloor n/2 \rfloor} \binom{n}{2k} \binom{k}{j} = 2^{n-2j-1} \binom{n-j}{j} \frac{n}{n-j}. \)

1.121

\(\sum_{k=j}^{n} \binom{n+1}{2k+1} \binom{k}{j} = 2^{n-2j} \binom{n-j}{j}. \)

1.122

\(\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \binom{n+k-1}{k} \frac{1}{k} = 2 \sum_{k=1}^{n} \frac{1}{k}. \)

1.123

\(\sum_{k=1}^{n} (-1)^k \binom{n}{k} \binom{n+k-1}{k} \sum_{j=1}^{k} \frac{1}{j} = \frac{(-1)^n}{n}. \)

1.124 R.R. Goldberg

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{n+k-1}{k} \sum_{j=1}^{n+k-1} \frac{1}{j} = \frac{(-1)^n}{n} \)

1.125

\(\sum_{k=1}^{n} \binom{n}{k}^2 \sum_{j=1}^{k} \frac{1}{j} = \binom{2n}{n} \left( \sum_{j=1}^{n} \frac{1}{j} - \sum_{j=n+1}^{2n} \frac{1}{j} \right). \)

1.126

\(\sum_{k=0}^{\infty} \binom{1/2}{k}^2 \frac{1}{2n+2k} = \sum_{k=0}^{\infty} \binom{1/2}{k}^2 \frac{1}{2n-2k+1}, \quad (n \geq 1). \)

1.127

\(\sum_{k=0}^{\infty} \binom{2k}{k}^2 \left(\frac{z}{4}\right)^{2k} = \frac{2}{\pi} \int_{0}^{\pi/2} \frac{dx}{\sqrt{1 - z^2 \sin^2 x}}. \)

1.128

\(\sum_{k=0}^{\infty} \binom{2k}{k}^2 \left(\frac{z}{4}\right)^{2k} \frac{1}{1-2k} = \frac{2}{\pi} \int_{0}^{\pi/2} \sqrt{1 - z^2 \sin^2 x} \, dx. \)

Section 2

1.1 Binomial Theorem

\(\sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k} = (x + y)^n\)

1.2 Alternating sum of Binomial coefficients to infinity is 0

\(\sum_{k=0}^{\infty} (-1)^k \binom{x}{k} = 0\)

For a constant x, and k is a variable between 0 and infinity.

Citation : MathStackExchange

1.3 Sum of a polynomial and binomial coefficient is a rational function

\(\sum_{k=0}^{\infty} \binom{n+k}{k} x^k = \frac{1}{(1 - x)^{n+1}}, \quad |x| < 1\)

1.4 Sum of a constant alternating coefficient is a two-sum of alternating coefficients.

\(\sum_{k=a}^{n} (-1)^k \binom{x}{k} = (-1)^a \binom{x-1}{a-1} + (-1)^n \binom{x-1}{n}\)

1.5 Finite sum of a constant alternating coefficient takes the sign of the final term in the series.

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} = (-1)^n \binom{x-1}{n}\)

1.5.1

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} = \prod_{k=1}^{n} \left( 1 - \frac{x}{k} \right),\)

1.5.2

\((-1)^n \binom{x-1}{n} = \prod_{k=1}^{n} \left( 1 - \frac{x}{k} \right),\)

1.6

\(\sum_{k=0}^{n} (-1)^k \binom{x}{k} k^r = \sum_{k=0}^{r} (-1)^k \binom{x}{k} \binom{n-x}{n-k} B_{k}^{r}\)

B is defined in [0.1] Sterling number of the second kind

1.7 An asymptotic relationship exists between alternating sums and the Gamm function.

\(\lim_{n \to \infty} \sum_{k=0}^{n-1} (-1)^k \binom{x}{k} k^r = \frac{1}{(r - x) \Gamma(-x)}\)

Where Γ is the gamma function and Γ(-x) = (-x−1)! for whole numbers

Source : American.Math.Monthly , 1953 p.482

1.8

\(\sum_{k=0}^{n} \binom{x+1}{k} z^k = \sum_{k=0}^{n} \binom{x - n + k}{k} z^k (1+z)^{n-k},\)

Where x and z are arbitrary constants.

1.9

\(\sum_{k=0}^{n} \binom{x}{k} y^k = \sum_{k=0}^{n} \binom{n-x}{k} (1+y)^{n-k} (-y)^k\)

1.10

\(\sum_{k=0}^{n-1} \binom{z}{k} x^{n-k-1} = \sum_{k=1}^{n} \binom{z-k}{n-k} (x+1)^{k-1} \)

1.11

\(\sum_{k=0}^{n-1} \binom{z}{k} \frac{x^{n-k}}{n-k} = \sum_{k=1}^{n} \binom{z-k}{n-k} \frac{(x+1)^k - 1}{k}\)

1.12

\(\sum_{k=0}^{n} \binom{n}{k} \frac{x^r}{r+k} = \sum_{k=1}^{r} (-1)^{r-k} \binom{r-1}{r-k} \frac{(x+1)^{n+k} - 1}{n+k}, \quad (r \geq 1)\)

1.13

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} k^j = \begin{cases} 0, & 0 \leq j < n \\ (-1)^n n!, & j = n \end{cases} \)

1.14

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} (x-k)^{n+1} = \frac{2x \cdot n}{2} (n+1)! \)

1.15

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} (x-k)^{n+2} = \frac{3n^2 + n + 12x^2 - 12nx}{24} (n+2)! \)

1.16

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} (x-k)^{n+j} = \sum_{k=0}^{j} \binom{x-n}{k} B_{n+k}^{n+j} \)

1.23

\(\sum_{k=0}^{\infty} \binom{n+k}{k} 2^{-k} = 2^{n+1} \)

1.24

\(\sum_{k=0}^{n} \binom{n}{k} = 2^n \)

1.25

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} = \begin{cases} 0, & n \neq 0 \\ 1, & n = 0 \end{cases} \)

1.26

\(\sum_{k=0}^{n} \binom{n}{k} \cos kx = 2^n \cos \frac{nx}{2} \left( \cos \frac{x}{2} \right)^n \)

1.27

\(\sum_{k=0}^{n} \binom{n}{k} \sin kx = 2^n \sin \frac{nx}{2} \left( \cos \frac{x}{2} \right)^n \)

1.28

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \cos kx = (-1)^n \cdot 2^n \left( \sin \frac{x}{2} \right)^n \cos \frac{n(x+\pi)}{2} \)

1.29

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \sin kx = (-1)^n \cdot 2^n \left( \sin \frac{x}{2} \right)^n \sin \frac{n(x+\pi)}{2} \)

1.30

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{k} \cos(n-2k)x = 2^{n-1} \cos^n x + \frac{1}{2} \binom{n}{n/2} \frac{1+(-1)^n}{2} \)

1.31

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \cos kx = 2^{n-1} \left\{ \cos^n \left(\frac{x}{4}\right) \cos \frac{nx}{4} + (-1)^n \sin^n \left(\frac{x}{4}\right) \cos \frac{n(2\pi - x)}{4} \right\} \)

1.32

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \sin kx = 2^{n-1} \left\{ \cos^n \left(\frac{x}{4}\right) \sin \frac{nx}{4} + (-1)^n \sin^n \left(\frac{x}{4}\right) \sin \frac{n(2\pi - x)}{4} \right\} \)

1.33

\(\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} \cos (2k+1)x = \\ 2^{n-1} \left\{ \cos^n (x/2) \cos (nx/2) - (-1)^n \sin^n (x/2) \cos \frac{n(\pi + x)}{2} \right\} \)

1.34

\(\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} \sin (2k+1)x = 2^{n-1} \left\{ \cos^n (x/2) \sin (nx/2) - (-1)^n \sin^n (x/2) \sin \frac{n(\pi + x)}{2} \right\} \)

1.35

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{k} = 2^{n-1} + \frac{1}{2} \binom{n}{n/2} \frac{1+(-1)^n}{2} \)

1.36

\(\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{k} = 2^{n-1} - \frac{1}{4} (-1)^n \binom{n}{n/2} \)

1.37

\(\sum_{k=0}^{n} \binom{n}{k} \frac{x^k}{k+1} = \frac{(x+1)^{n+1} - 1}{(n+1)x} \)

1.38

\(\sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} \frac{x^{2k}}{2k+1} = \frac{(x+1)^{n+1} - (1-x)^{n+1}}{2(n+1)x} \)

1.39

\(\sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} \frac{x^{2k}}{k+1} = \frac{(x+1)^{n+1} + (1-x)^{n+1} - 2}{(n+1)x^2} \)

1.40

\(\sum_{k=0}^{\infty} (-1)^k \binom{x}{k} \frac{1}{z+k} = \frac{\Gamma(z) \Gamma(x+1)}{\Gamma(x+z+1)} = \frac{1}{z} \binom{x+z}{x}, \quad R(x) > -1 \)

1.41

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x}{x+k} = \frac{1}{\binom{x+n}{n}} \)

1.42

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{(x+k)k} = \frac{x}{x+n} \)

1.43

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{x-k} = \frac{(-1)^n}{(x-n)} \binom{x}{n} \)

1.44

\(\sum_{k=0}^{n} (-1)^k \binom{n+1}{k+1} \frac{1}{k+1/a} = \sum_{k=0}^{n} \frac{1}{ak+1} \)

1.45

\(\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k} = \sum_{k=1}^{n} \frac{1}{k} \)

1.46

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{(k+3)(k+4)(k+7)} \)

=

\(= \frac{1}{n(n+1)} \left\{ \frac{1}{2(n+3)} + \frac{1}{2\binom{n+7}{5}} - \frac{1}{\binom{n+4}{2}} \right\} \)

1.47

\(\sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^k}{x+k} = (-1)^j \frac{x^{j-1}}{\binom{x+n}{n}}, \quad (j \leq n) \)

1.48

\(\sum_{k=0}^{n} \binom{k+x}{r} = \binom{n+x+1}{r+1} - \binom{x}{r+1} \)

1.49

\(\sum_{k=0}^{n} \binom{x+k}{k} = \binom{x+n+1}{n} \)

1.50

\(\sum_{k=1}^{n} \binom{x+k}{k} \frac{1}{x+k} = \frac{1}{x} \left\{ \binom{x+n}{n} - 1 \right\} \)

1.51

\(\sum_{k=a}^{n} \binom{k}{a} = \binom{n+1}{x+1} - \binom{a}{x+1} \)

1.52

\(\sum_{k=j}^{n} \binom{k}{j} = \binom{n+1}{j+1} \)

1.53

\(\sum_{k=0}^{\lfloor \frac{n-a}{r} \rfloor} \binom{n}{a+kr} x^{a+kr} = \frac{1}{r} \sum_{j=1}^{r} (\omega_r^j)^{-a} (1 + x \omega_r^j)^n \)

where

\(\omega_r = e^{2\pi i / r} \)

and

\((r-1 \geq a), \quad (a = \text{positive or negative integer}) \)

1.54

\(\sum_{k=0}^{\lfloor \frac{n-a}{r} \rfloor} \binom{n}{a+kr} = \frac{1}{r} \sum_{j=1}^{r} \left( 2 \cos \frac{\pi j}{r} \right)^n \cos \frac{(n-2a)j\pi}{r} \)

where

\((n \geq a \geq 0, r-1 \geq a) \)

1.55

\(\sum_{k=0}^{\lfloor \frac{n}{r} \rfloor} \binom{n}{rk} = \frac{1}{r} \sum_{j=1}^{r} \left( 1 + \omega_r^j \right)^n \)

which simplifies to

\(= \frac{2}{r} \sum_{j=1}^{r} \left( \cos \frac{\pi j}{r} \right)^n \cos \frac{n \pi j}{r} \)

1.56

\(\sum_{k=0}^{\lfloor \frac{n}{3} \rfloor} \binom{n}{3k} = \frac{1}{3} \left\{ 2^n + 2 \cos \frac{n \pi}{3} \right\} \)

1.57

\(\sum_{k=0}^{\lfloor \frac{n-1}{3} \rfloor} \binom{n}{3k+1} = \frac{1}{3} \left\{ 2^n + 2 \cos \frac{(n-2)\pi}{3} \right\} \)

1.58

\(\sum_{k=0}^{\lfloor \frac{n}{4} \rfloor} \binom{n}{4k} = \frac{1}{4} \left\{ 2^n + 2(\sqrt{2})^n \cos \frac{n\pi}{4} \right\} \)

1.59

\(\sum_{k=0}^{n} \binom{4n}{4k} = \frac{1}{4} \left\{ 2^{4n} + (-1)^n 2^{2n+1} \right\} \)

1.60

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} (xy)^k (x+y)^{n-2k} = \frac{x^{n+1} - y^{n+1}}{x - y} \)

1.61

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} 2^{n-2k} z^k = \frac{x^{n+1} - y^{n+1}}{x - y} \)

where

\(x = 1 + \sqrt{z+1}, \quad y = 1 - \sqrt{z+1} \)

1.62

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} (2 \cos x)^{n-2k} = \frac{\sin (n+1)x}{\sin x} \)

1.63

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} \frac{(2 \cos x)^{n-2k}}{n-k} = \frac{2}{n} \cos nx \)

1.64

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} \frac{1}{n-k} \left(\frac{z}{4}\right)^k = \frac{1}{n 2^{n-1}} \frac{x^n - y^n}{x+y} \)

where

\(x = 1+\sqrt{z+1}, \quad y = 1 - \sqrt{z+1} \)

1.65

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} \frac{2^{2n-2k}}{n-k} = \frac{2^n + 1}{n}, \quad (n > 1) \)

1.66

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} \frac{(2 \cos x)^{n-2k}}{k+1} = \frac{(2 \cos x)^{n+2} - 2 \cos(n+2)x}{n+2} \)

1.67

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} \frac{2^{2n-2k}}{4(k+1)} = \frac{n+1}{n+2} - \frac{2^n + 1}{n+2} \)

1.68

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} \frac{1}{n-k} = \begin{cases} (-1)^n \frac{2}{n}, & \text{if } 3 \mid n, \\ (-1)^{n-1} \frac{1}{n}, & \text{if } 3 \nmid n. \end{cases} \)

source : ( Cambridge Math. Tripos, 1932; Hardy, Pure Math., page 445.)

1.69

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} \frac{k}{n-k} 6^k = \frac{3^n + (-1)^n 2^n}{n}, \quad (n > 1) \)

1.70

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} z^k = \frac{\left(2z + 1 \pm \sqrt{1+4z} \right)^{\frac{n+1}{2}} + (-1)^n (2z)^{\frac{n+1}{2}}} {2 \left(2z + 1 \pm \sqrt{1+4z} \right)^{\frac{n}{2}} (4z + 1 \pm \sqrt{1+4z})} \)

1.71

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} z^k = \frac{x^{n+1} - 1}{(x - 1)(1+x)^n}, \quad \text{where: } z = \frac{-x^2}{(1+x)^2} \)

1.72

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} 2^{n-2k} = n+1 \)

1.73

\(\sum_{k=0}^{n} (-1)^k \binom{2n-k}{k} \frac{1}{2^{2k}} = \frac{2n+1}{2^{2n}}, \)

1.74

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n-k}{k} = \frac{(1+\sqrt{5})^{n+1} - (1-\sqrt{5})^{n+1}}{2^{n+1} \sqrt{5}} \)

where an​ is the n-th Fibonacci number, defined by:

\(a_0 = 1, \quad a_1 = 1, \quad a_{n+1} = a_n + a_{n-1}. \)

1.75

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n-k}{k} = \frac{(-1)^{\lfloor \frac{n}{3} \rfloor} + (-1)^{\lfloor \frac{n+1}{3} \rfloor}}{2}, \)

1.76

\(\sum_{k=0}^{n} \binom{n+k}{2k} = \sum_{k=0}^{n} \binom{2n-k}{k} = a_{2n}, \quad \text{using (1.74)}, \)

1.77

\(\sum_{k=0}^{n} \binom{n+k}{2k} 2^{n-k} = \frac{2^{2n+1} + 1}{3}, \)

1.78

\(\sum_{k=0}^{n} \binom{n+k}{k} \left\{ (1-x)^{n+1} x^k + x^{n+1} (1-x)^k \right\} = 1, \)

1.79

\(\sum_{k=0}^{n} \binom{n+k}{k} 2^{-k} = 2^n, \)

1.80

\(\sum_{k=1}^{\infty} \binom{2n+k}{n} 2^{-k} = 2^{2n}, \)

1.81

\(\sum_{k=0}^{n} \binom{2n-k}{n} 2^k = 2^{2n}, \)

1.82

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{2n}{n+2k} = 2^{2n-2} + \binom{2n-1}{n}, \quad (n \geq 1), \)

1.83

\(\sum_{k=0}^{n} \binom{2n+1}{k} = 2^{2n}, \)

1.84

\(\sum_{k=0}^{n} \binom{2n-1}{k} = 2^{2n-2} + \binom{2n-1}{n}, \quad (n \geq 1), \)

1.85

\(\sum_{k=0}^{n} \binom{2n}{k} = 2^{2n-1} + \binom{2n-1}{n}, \quad (n \geq 1), \)

1.86

\(\sum_{k=0}^{n} (-1)^k \binom{2n}{k} = (-1)^n \binom{2n-1}{n}, \)

1.87

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} x^k = \frac{(1+\sqrt{x})^n + (1-\sqrt{x})^n}{2}, \)

1.88

\(\sum_{k=0}^{n} (-1)^k \binom{2n}{2k} = (-1)^{n/2} \binom{n}{2} \frac{1+(-1)^n}{2}, \)

1.89

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = 2^{n-1}, \quad (n \geq 1), \)

1.90

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \binom{n}{2k} = \frac{(1+1)^n + (1-1)^n}{2} = (\sqrt{2})^n \cos \frac{n\pi}{4}, \)

1.91

\(\sum_{k=0}^{n} \binom{2n}{2k} = \sum_{k=0}^{n} \binom{2n}{2k+1} = 2^{2n-1}, \quad (n \geq 1), \)

1.92

\(\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{2n}{2k} = 2^{2n-2} + \frac{1+(-1)^n}{2} \binom{2n-1}{n}, \quad (n \geq 1), \)

1.93

\(\sum_{k=0}^{n} \binom{2n+1}{2k} = \sum_{k=0}^{n} \binom{2n+1}{2k+1} = 2^{2n}, \)

1.94

\(\sum_{k=0}^{n} (-1)^k \binom{2n+1}{2k} = (-1)^{\lfloor \frac{n+1}{2} \rfloor} 2^n, \)

1.95

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{2k+1} x^k = \frac{(1+\sqrt{x})^n - (1-\sqrt{x})^n}{2\sqrt{x}}, \quad (n \geq 1), \)

1.96

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} (-1)^k \binom{n}{2k+1} = \frac{(1+1)^n - (1-1)^n}{2} = (\sqrt{2})^n \sin \frac{n\pi}{4}, \)

1.97

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{n}{2k+1} = 2^{n-1}, \)

1.98

\(\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} \binom{2n}{2k+1} = 2^{2n-2} + \frac{1-(-1)^n}{2} \binom{2n-1}{n}, \)

1.99

\(\sum_{k=0}^{n-1} (-1)^k \binom{2n}{2k+1} = (-1)^{\lfloor \frac{n}{2} \rfloor} 2^n \frac{1 - (-1)^n}{2}, \)

1.100

\(\sum_{k=0}^{n} \binom{2n+1}{2k+1} k = (2n-1) 2^{2n-2}, \quad (n \geq 1), \)

1.101

\(\sum_{k=0}^{n} (-1)^k \binom{2n+1}{2k+1} = (-1)^{\lfloor \frac{n}{2} \rfloor} 2^n, \)

1.102

\(\sum_{k=0}^{n} (-1)^k \binom{n+x}{n-k} \frac{1}{k+1} = \frac{B^n(x) + n}{n}, \quad \text{symbolically}, \)

where Bⁿ (x) represents the Bernoulli polynomial.

1.103

\(\sum_{k=0}^{n} \binom{n}{k} \frac{k \cdot k!}{(n+1) x^{n+1}} = 1 - \frac{x (n+1)!}{(n+1)!}, \)

1.104

\(\sum_{k=0}^{\infty} \binom{2k}{k} \frac{x^k}{(1 - 2k) 2^{2k}} = \sqrt{1 - x}, \quad (|x| \leq 1), \)

1.105

\(\sum_{k=0}^{\infty} \binom{2k}{k} \frac{x^k}{2^{2k}} = \frac{1}{\sqrt{1 - x}}, \quad (|x| < 1, \text{ also valid for } x = -1), \)

1.106

\(\sum_{k=1}^{\infty} (-1)^{k+1} \binom{2k}{k} \frac{x^k}{k 2^{2k+1}} = \log \frac{1 + \sqrt{1 + x}}{2}, \quad (|x| < 1), \)

1.107

\(\sum_{k=0}^{\infty} \binom{2k}{k} \frac{1}{(2k+1)^2 2^{2k}} = \int_{0}^{1} \frac{\arcsin x}{x} \, dx = \frac{\pi}{2} \log 2. \)

1.108

\(\sum_{k=0}^{n} \binom{2k}{k} \frac{k \cdot k!}{2^{2k}} s_r = \frac{2n+1}{2n} s_r = \binom{2n}{n} \sum_{k=0}^{n} \binom{n}{k} B_k^{r} 2k+1, \)

1.109

\(s_0 = \frac{2n+1}{2n} \binom{2n}{n} = \binom{n + \frac{1}{2}}{n}, \)

1.110

\(s_1 = \frac{n}{3} s_0, \)

1.111

\(s_2 = \frac{n(3n+2)}{3(5)} s_0, \)

1.112

\(s_3 = \frac{n(15n^2 + 18n + 2)}{3(5)(7)} s_0, \)

1.113

\(\sum_{k=0}^{n} \binom{2k}{k} \frac{k \cdot k!}{2^k} = \binom{2n+2}{n+1} \frac{(n+1)!}{2^{n+2}} - \frac{1}{2}, \)

1.114

\(B_n^f (x) = \sum_{k=0}^{n} \binom{n}{k} (1-x)^{n-k} x^k f \left( \frac{k}{n} \right), \)

1.115

\(L_n^{(a)} (x) = \sum_{k=0}^{n} (-1)^k \binom{n+a}{n-k} \frac{x^k}{k!}, \)

\(= \frac{1}{n!} x^{-a} e^x D_x^n \left\{ x^{a+n} e^{-x} \right\}, \)

\(\text{Ordinary Laguerre Polynomials: } L_n (x) = L_n^{(0)} (x). \)

1.116

\(\sum_{k=0}^{n} \binom{p-k}{n-k} x^k = x^{p+1} (x-1)^{n-p-1}, \quad \text{provided } 0 \leq p \leq n-1. \)

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Footnotes

1. Graham, R. L., Knuth, D. E., & Patashnik, O. (1994). Concrete mathematics: A foundation for computer science (2nd ed.). Addison-Wesley.

2. Riordan, J. (1968). Combinatorial identities. Wiley.

3. Gould, H. W. (1972). Combinatorial identities: A standardized set of tables listing 500 binomial coefficient summations. Morgantown Printing & Binding Co.

4. OEIS Foundation Inc. (2025). The On-Line Encyclopedia of Integer Sequences. https://oeis.org