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# On Alternating Sums of Binomial Coefficients

Soham Dutta

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Soham Dutta. On Alternating Sums of Binomial Coefficients. 2024. hal-04580266v2

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# On Alternating Sums of Binomial Coefficients

Soham Dutta

► To cite this version:

Soham Dutta. On Alternating Sums of Binomial Coefficients. 2024. hal-04580266v2

HAL Id: hal-04580266

https://hal.science/hal-04580266v2

Preprint submitted on 20 Jun 2024

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
ON ALTERNATING SUMS OF BINOMIAL
COEFFICIENTS

SOHAM DUTTA

UNDERGRADUATE STUDENT

INDIAN STATISTICAL INSTITUTE, KOLKATA

EMAIL ID: BS2347@ISICAL.AC.IN

Date: May 2024

Modified: June 2024.
ON ALTERNATING SUMS OF BINOMIAL
COEFFICIENTS

SOHAM DUTTA

UNDERGRADUATE STUDENT

INDIAN STATISTICAL INSTITUTE, KOLKATA

EMAIL ID: BS2347@ISICAL.AC.IN

Date: May 2024

Modified: June 2024.
ABSTRACT. By considering the binomial expansion formula of $(j-x)^n$,
we provide a method that allows us to come up with infinitely many
non-trivial identities involving alternating sums of binomial coefficients.
This approach is further extended by using Beta and Incomplete Beta
Functions. Towards the end, we also explore an invariance in the iden-
tities and obtain some general limits (in terms of $n$).

AMS Subject Classification:33B15, 33B20, 26A36, 26A99

Keywords: Beta Function, Incomplete Beta Function, Integral Representations
ABSTRACT. By considering the binomial expansion formula of $(j-x)^n$,
we provide a method that allows us to come up with infinitely many
non-trivial identities involving alternating sums of binomial coefficients.
This approach is further extended by using Beta and Incomplete Beta
Functions. Towards the end, we also explore an invariance in the iden-
tities and obtain some general limits (in terms of $n$).

AMS Subject Classification:33B15, 33B20, 26A36, 26A99

Keywords: Beta Function, Incomplete Beta Function, Integral Representations
# 1. INTRODUCTION

The binomial theorem states that

$$ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} $$

Putting $(a, b) = (1, x)$ and $(1, -x)$ we have the following two identities respectively

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

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

Some common binomial identities which are pretty direct from the above are

$$ \sum_{k=0}^{n} \binom{n}{k} = 2^n $$

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

The first one follows from putting $x = 1$ in (1) and the second one follows from putting $x = 1$ in (2).

Binomial identities have been a very interesting field of research for long. The earliest occurrence of research involving binomial coefficients dates back to the time of Pascal (which was published posthumously in 1655). Since then, a lot of interesting research has been done. Some notable works include those by Fine (1947), Knuth and Wilf (1989), Sofo (2009), Sun (2010). In 2004, Sury [4] discovered a set of identities involving reciprocal of binomial coefficients. Batir and Sofo (2021) [1] also did significant work on deriving identities involving reciprocals of binomial coefficients.

In this paper, we exhibit a method to generate lots of identities related to alternating sums of binomial coefficients. The main stepping stone for our “method” is using Beta [5] [2] and Incomplete Beta Functions [6] [3] [2]. Co-incidentally, we also obtain a (pseudo) recursive relation for the Beta and Incomplete Beta Function.
# 1. INTRODUCTION

The binomial theorem states that

$$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} $$

Putting $(a,b) = (1,x)$ and $(1,-x)$ we have the following two identities respectively

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

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

Some common binomial identities which are pretty direct from the above are

$$ \sum_{k=0}^{n} \binom{n}{k} = 2^n $$

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

The first one follows from putting $x = 1$ in (1) and the second one follows from putting $x = 1$ in (2).

Binomial identities have been a very interesting field of research for long. The earliest occurrence of research involving binomial coefficients dates back to the time of Pascal (which was published posthumously in 1655). Since then, a lot of interesting research has been done. Some notable works include those by Fine (1947), Knuth and Wilf (1989), Sofo (2009), Sun (2010). In 2004, Sury [4] discovered a set of identities involving reciprocal of binomial coefficients. Batir and Sofo (2021) [1] also did significant work on deriving identities involving reciprocals of binomial coefficients.

In this paper, we exhibit a method to generate lots of identities related to alternating sums of binomial coefficients. The main stepping stone for our "method" is using Beta [5] [2] and Incomplete Beta Functions [6] [3] [2]. Co-incidentally, we also obtain a (pseudo) recursive relation for the Beta and Incomplete Beta Function.
The results in this paper are involving real numbers. However, they can
easily be extended to complex numbers too.

## 2. THE METHOD

We again start with the binomial expansion of $(1-x)^n$. We have that

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

Consider the integral of the above expression over $[0, t]$ (where $t \le 1$).
Replacing the final $t$ by $x$, we have

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

Multiplying both sides of the expression by $x^m$ where $m \in \mathbb{R} \setminus \{-1\}$, we
have

$$ \frac{-(1-x)^{n+1}x^m}{n+1} + \frac{x^m}{n+1} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+k+1}}{k+1} \quad (3) $$

Now integrating both sides of the expression from 0 to 1, we have that

$$ \begin{aligned} \frac{-1}{n+1} \int_{0}^{1} x^m (1-x)^{n+1} dx + \frac{1}{n+1} \int_{0}^{1} x^m dx &= \int_{0}^{1} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+k+1}}{k+1} dx \\ &\implies \frac{-1}{n+1} \times \beta(m+1, n+2) + \frac{1}{(m+1)(n+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{(k+1)(m+k+2)} \end{aligned} $$

Using the fact that $\beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$, we finally have that

$$ \frac{1}{(m+1)(n+1)} - \frac{\Gamma(m+1)\Gamma(n+2)}{\Gamma(m+n+3) \times (n+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{(k+1)((k+1)+(m+1))} \quad (4) $$

Now instead of integrating (3) from 0 to 1, if one integrates from 0 to $t$
where $t \in [0, 1]$, we have

$$ \frac{-1}{n+1} \int_{0}^{t} x^m (1-x)^{n+1} dx + \frac{1}{n+1} \int_{0}^{t} x^m dx = \int_{0}^{t} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{k+m+1}}{k+1} \quad (4) $$
The results in this paper are involving real numbers. However, they can
easily be extended to complex numbers too.

## 2. THE METHOD

We again start with the binomial expansion of $(1-x)^n$. We have that

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

Consider the integral of the above expression over $[0, t]$ (where $t \le 1$).
Replacing the final $t$ by $x$, we have

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

Multiplying both sides of the expression by $x^m$ where $m \in \mathbb{R} \setminus \{-1\}$, we
have

$$ \frac{-(1-x)^{n+1}x^m}{n+1} + \frac{x^m}{n+1} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+k+1}}{k+1} \quad (3) $$

Now integrating both sides of the expression from 0 to 1, we have that

$$ \begin{aligned} & \frac{-1}{n+1} \int_{0}^{1} x^m (1-x)^{n+1} dx + \frac{1}{n+1} \int_{0}^{1} x^m dx = \int_{0}^{1} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+k+1}}{k+1} \\ & \implies \frac{-1}{n+1} \times \beta(m+1, n+2) + \frac{1}{(m+1)(n+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{(k+1)(m+k+2)} \end{aligned} $$

Using the fact that $\beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$, we finally have that

$$ \frac{1}{(m+1)(n+1)} - \frac{\Gamma(m+1)\Gamma(n+2)}{\Gamma(m+n+3) \times (n+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{1}{(k+1)((k+1)+(m+1))} \quad (4) $$

Now instead of integrating (3) from 0 to 1, if one integrates from 0 to $t$
where $t \in [0, 1]$, we have

$$ \frac{-1}{n+1} \int_{0}^{t} x^m (1-x)^{n+1} dx + \frac{1}{n+1} \int_{0}^{t} x^m dx = \int_{0}^{t} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{k+m+1}}{k+1} \qquad 4 $$
Evaluating the above expression using the definition of the Incomplete Beta Function, we have the following identity (after replacing the final t by x)

$$ \left(\frac{-1}{n+1}\right) \times \beta(x; m+1, n+2) + \frac{x^{m+1}}{(m+1)(n+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+k+2}}{(k+1)(m+k+2)} \quad (5) $$

### 3. GENERALIZATION OF THE METHOD (TYPE 1)

In the above method, if one starts with $(j-x)^n$ instead of $(1-x)^n$ (where $j \in \mathbb{R} \setminus \{0\}$), continuing in the same way as above, one would arrive at the following

$$ \frac{-(j-x)^{n+1}x^m}{n+1} + \frac{j \times x^m}{n+1} = \sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k} \times j^{n-k} x^{k+m+1}}{k+1} \quad (6) $$

Now integrating this expression from 0 to 1, we have

$$ \int_0^1 \frac{-(j-x)^{n+1}x^m}{n+1} dx + \int_0^1 \frac{j \times x^m}{n+1} dx = \int_0^1 \sum_{k=0}^n (-1)^k \frac{\binom{n}{k} \times j^{n-k} x^{k+m+1}}{k+1} dx $$

Using the definition of the Incomplete Beta Function, we can evaluate the above expression as follows

$$ \frac{j}{(n+1)(m+1)} - \frac{j^{m+n+2}}{n+1} \times \beta\left(\frac{1}{j}; m+1, n+2\right) = \sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k} j^{n-k}}{(k+m+2)(k+1)} \quad (7) $$

### 4. GENERALIZATION OF THE METHOD (TYPE 2)

Consider (5). Multiplying both sides of the equality by $x^p$ ($p \in \mathbb{R} \setminus \{-1\}$) and integrating from 0 to t ($t \in [0, 1]$), we have

$$ \begin{aligned} & \frac{-1}{n+1} \int_0^t \beta(x; m+1, n+2) x^p dx + \int_0^t \frac{x^{m+p+1}}{(m+1)(n+1)} dx \\ & = \int_0^t \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{x^{m+p+k+2}}{(k+1)((m+1)+(k+1))} dx \end{aligned} \quad (8) $$
Evaluating the above expression using the definition of the Incomplete Beta Function, we have the following identity (after replacing the final $t$ by $x$)

$$ \left(\frac{-1}{n+1}\right) \times \beta(x; m+1, n+2) + \frac{x^{m+1}}{(m+1)(n+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+k+2}}{(k+1)(m+k+2)} \quad (5) $$

### 3. GENERALIZATION OF THE METHOD (TYPE 1)

In the above method, if one starts with $(j-x)^n$ instead of $(1-x)^n$ (where $j \in \mathbb{R}\setminus\{0\}$), continuing in the same way as above, one would arrive at the following

$$ \frac{-(j-x)^{n+1}x^m}{n+1} + \frac{j \times x^m}{n+1} = \sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k} \times j^{n-k} x^{k+m+1}}{k+1} \quad (6) $$

Now integrating this expression from 0 to 1, we have

$$ \int_0^1 \frac{-(j-x)^{n+1}x^m}{n+1} dx + \int_0^1 \frac{j \times x^m}{n+1} dx = \int_0^1 \sum_{k=0}^n (-1)^k \frac{\binom{n}{k} \times j^{n-k} x^{k+m+1}}{k+1} dx $$

Using the definition of the Incomplete Beta Function, we can evaluate the above expression as follows

$$ \frac{j}{(n+1)(m+1)} - \frac{j^{m+n+2}}{n+1} \times \beta\left(\frac{1}{j}; m+1, n+2\right) = \sum_{k=0}^{n} (-1)^k \frac{\binom{n}{k} j^{n-k}}{(k+m+2)(k+1)} \quad (7) $$

### 4. GENERALIZATION OF THE METHOD (TYPE 2)

Consider (5). Multiplying both sides of the equality by $x^p$ ($p \in \mathbb{R}\setminus\{-1\}$) and integrating from 0 to $t$ ($t \in [0, 1]$), we have

$$ \begin{aligned} & \frac{-1}{n+1} \int_0^t \beta(x; m+1, n+2) x^p dx + \int_0^t \frac{x^{m+p+1}}{(m+1)(n+1)} dx \\ & = \int_0^t \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{x^{m+p+k+2}}{(k+1)((m+1)+(k+1))} dx \end{aligned} \quad (8) $$
On complete evalution of (8) and replacing the final $t$ by $x$, we arrive at the following equality

$$ \frac{x^{m+p+2}}{(m+1)(n+1)(m+p+2)} - \frac{1}{n+1}\beta(x; m+1, n+2)\frac{x^{p+1}}{p+1} - \frac{1}{p+1}\beta(x; m+p+2, n+2) \\ = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+p+k+3}}{(m+p+k+3)(m+k+2)(k+1)} \quad (9) $$

### 5. GENERALIZATION OF THE METHOD (TYPE 3)

Consider the expression in (6) which is

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

Instead of integrating it from 0 to 1, if we integrate it from 0 to $t$ where $t \in [0, 1]$, we have

$$ \int_{0}^{t} \frac{-(j-x)^{n+1}x^m}{n+1} dx + \int_{0}^{t} \frac{j \times x^m}{n+1} dx = \int_{0}^{t} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^{n-k}x^{k+m+1}}{k+1} dx $$

This evaluates to (after evaluating the final $t$ by $x$)

$$ \beta\left(\frac{x}{j}; m+1, n+2\right) \times \left(\frac{-1}{n+1}\right) + \frac{j \times x^{m+1}}{(n+1)(m+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^{n-k} \times x^{k+m+2}}{(k+1)(k+m+2)} \quad (10) $$

Now multiplying both sides of the equation by $x^p$ ($p \in \mathbb{R} \setminus \{-1\}$), integrating from 0 to $t$ ($t \in [0, 1]$) and replacing the final $t$ by $x$, one obtains the following

$$ \frac{j^{p+1}}{(p+1)(n+1)} \beta\left(\frac{x}{j}; m+p+2, n+2\right) - \beta\left(\frac{x}{j}; m+1, n+2\right) \frac{x^{p+1}}{(p+1)(n+1)} \\ + \frac{j \times x^{m+p+2}}{(n+1)(m+1)(m+p+2)} \\ = \\ \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^{n-k}}{(k+1)(k+m+2)} \cdot \frac{x^{m+k+p+3}}{(m+k+p+3)} \quad (11) $$
On complete evalution of (8) and replacing the final $t$ by $x$, we arrive at the following equality

$$ \frac{x^{m+p+2}}{(m+1)(n+1)(m+p+2)} - \frac{1}{n+1}\beta(x; m+1, n+2)\frac{x^{p+1}}{p+1} - \frac{1}{p+1}\beta(x; m+p+2, n+2) \\ = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{x^{m+p+k+3}}{(m+p+k+3)(m+k+2)(k+1)} \quad (9) $$

### 5. GENERALIZATION OF THE METHOD (TYPE 3)

Consider the expression in (6) which is

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

Instead of integrating it from 0 to 1, if we integrate it from 0 to $t$ where $t \in [0, 1]$, we have

$$ \int_0^t \frac{-(j-x)^{n+1}x^m}{n+1} dx + \int_0^t \frac{j \times x^m}{n+1} dx = \int_0^t \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^{n-k} x^{k+m+1}}{k+1} dx $$

This evaluates to (after evaluating the final $t$ by $x$)

$$ \beta\left(\frac{x}{j}; m+1, n+2\right) \times \left(\frac{-1}{n+1}\right) + \frac{j \times x^{m+1}}{(n+1)(m+1)} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^{n-k} \times x^{k+m+2}}{(k+1)(k+m+2)} \quad (10) $$

Now multiplying both sides of the equation by $x^p$ ($p \in \mathbb{R} \setminus \{-1\}$), integrating from 0 to $t$ ($t \in [0, 1]$) and replacing the final $t$ by $x$, one obtains the following

$$ \frac{j^{p+1}}{(p+1)(n+1)} \beta\left(\frac{x}{j}; m+p+2, n+2\right) - \beta\left(\frac{x}{j}; m+1, n+2\right) \frac{x^{p+1}}{(p+1)(n+1)} \\ + \frac{j \times x^{m+p+2}}{(n+1)(m+1)(m+p+2)} \\ = \\ \sum_{k=0}^{n} (-1)^k \binom{n}{k} \frac{j^{n-k}}{(k+1)(k+m+2)} \cdot \frac{x^{m+k+p+3}}{(m+k+p+3)} \quad (11) $$
# 6. THE SPECIAL CASE INVOLVING $x^{-1}$

In our above discussion, we have avoided multiplying with $x^{-1}$. Multiplying by $x^{-1}$ and then integrating leads to the introduction of log terms in the formula which in turn make it difficult to extend. They can be extended by integrating by parts but the formulas obtained are way too clumsy and are presumably not of any real use.

# 7. SUMMARY

In this paper, we have given a method to generate infinitely many identities involving the binomial coefficients. We have derived several novel formulas namely (5), (7), (10), (11), (9). More formulas could have been derived but in order to keep this paper short, we have avoided the generalization of the methods, especially those mentioned as Type 2 and Type 3. It is interesting to note that in the formulas (7), (10), (11), (9), the terms in the LHS which are of the form $k\beta(x; a, b)x^q$ (where $q$ may be 0 also and $k$ is a constant) have the value of $a+q$ as an invariant. For example in (9), the said terms are $\beta(x; m+1, n+2)x^{p+1}$ and $\beta(x; m+p+2, n+2)x^0$ which have the required common value as $m+p+2$.

Also note that in all the formulas described in this paper, the alternating binomial sum present on the right hand side is $O(\frac{1}{n})$. By fixing $x, j$ and choosing $n, m, p, k$ as necessary, one can obtain some interesting limits of sums involving binomial coefficients.

# 8. FURTHER WORK

It will be interesting to note if there exists an absolute closed-form generalization of the formulas derived above. With this in mind, we propose the following questions for further research and exploration:

1. Does there exist a closed-form generalization in terms of $z$ of the Type 2 Method where the operator $x^{r_i} \int_0^t$ has been applied $z$ times?
(Here $r_i$ is a real number which is not $-1$, $t \in [0, 1]$ is a real and $z$ is a natural number.)
## 6. THE SPECIAL CASE INVOLVING $x^{-1}$

In our above discussion, we have avoided multiplying with $x^{-1}$. Multiplying by $x^{-1}$ and then integrating leads to the introduction of log terms in the formula which in turn make it difficult to extend. They can be extended by integrating by parts but the formulas obtained are way too clumsy and are presumably not of any real use.

## 7. SUMMARY

In this paper, we have given a method to generate infinitely many identities involving the binomial coefficients. We have derived several novel formulas namely (5), (7), (10), (11), (9). More formulas could have been derived but in order to keep this paper short, we have avoided the generalization of the methods, especially those mentioned as Type 2 and Type 3. It is interesting to note that in the formulas (7), (10), (11), (9), the terms in the LHS which are of the form $k\beta(x; a, b)x^q$ (where $q$ may be 0 also and $k$ is a constant) have the value of $a + q$ as an invariant. For example in (9), the said terms are $\beta(x; m + 1, n + 2)x^{p+1}$ and $\beta(x; m + p + 2, n + 2)x^0$ which have the required common value as $m + p + 2$.

Also note that in all the formulas described in this paper, the alternating binomial sum present on the right hand side is $O(\frac{1}{n})$. By fixing $x, j$ and choosing $n, m, p, k$ as necessary, one can obtain some interesting limits of sums involving binomial coefficients.

## 8. FURTHER WORK

It will be interesting to note if there exists an absolute closed-form generalization of the formulas derived above. With this in mind, we propose the following questions for further research and exploration:

(1) Does there exist a closed-form generalization in terms of $z$ of the Type 2 Method where the operator $x^{r_i} \int_0^t$ has been applied $z$ times?
(Here $r_i$ is a real number which is not $-1$, $t \in [0, 1]$ is a real and $z$ is a natural number.)
(2) Does there exist a closed-form generalization in terms of $z$ of the Type 3 Method where the operator $x^{r_i} \int_0^t$ has been applied $z$ times?
(Here $r_i$ is a real number which is not $-1$, $t \in [0, 1]$ is a real and $z$ is a natural number.)

## 9. ACKNOWLEDGEMENTS

I would like to thank Prof. Arnab Chakraborty of the Indian Statistical Institute, Kolkata for going through this paper and suggesting valuable changes. I would like to thank my parents and family for their constant support.

## 10. CONFLICT OF INTEREST

The author declares no conflict of interest.

## 11. FUNDING INFORMATION

There has been no funding for the work presented in this paper.

## 12. APPENDIX

The proofs of Theorems (7), (10), (11) and (9) which were omitted above are given in this supplementary section.

$$ \text{THE INTEGRAL } \int_{0}^{t} x^{m}(j-x)^{n} dx $$

Consider the integral $\int_{0}^{t} x^{m}(j-x)^{n} dx$ where $t \in [0, 1]$ and $j \in \mathbb{R} \setminus \{0\}$. Consider the change of variable given by $x \mapsto jx$. Then

$$ I_1 = \int_{0}^{t} x^{m}(j-x)^{n} dx = \int_{0}^{\frac{t}{j}} j^{m+n+1} x^{m}(1-x)^{n} dx = j^{m+n+1} \beta\left(\frac{t}{j}; m+1, n+1\right) $$

This readily yields the proof of (10). The proof of (7) also follows trivially once we set $t = 1$ in the last line of the equality.

$$ \text{THE INTEGRAL } \int_{0}^{t} x^{p} \cdot \beta\left(\frac{x}{j}; m, n\right) dx $$

Consider the integral

$$ I_2 = \int_{0}^{t} x^{p} \cdot \beta\left(\frac{x}{j}; m, n\right) dx $$
(2) Does there exist a closed-form generalization in terms of $z$ of the Type 3 Method where the operator $x^{r_i} \int_0^t$ has been applied $z$ times?
(Here $r_i$ is a real number which is not $-1$, $t \in [0, 1]$ is a real and $z$ is a natural number.)

## 9. ACKNOWLEDGEMENTS

I would like to thank Prof. Arnab Chakraborty of the Indian Statistical Institute, Kolkata for going through this paper and suggesting valuable changes. I would like to thank my parents and family for their constant support.

## 10. CONFLICT OF INTEREST

The author declares no conflict of interest.

## 11. FUNDING INFORMATION

There has been no funding for the work presented in this paper.

## 12. APPENDIX

The proofs of Theorems (7), (10), (11) and (9) which were omitted above are given in this supplementary section.

$$ \text{THE INTEGRAL } \int_{0}^{t} x^{m}(j-x)^{n} dx $$

Consider the integral $\int_{0}^{t} x^{m}(j-x)^{n} dx$ where $t \in [0, 1]$ and $j \in \mathbb{R} \setminus \{0\}$. Consider the change of variable given by $x \mapsto jx$. Then

$$ I_1 = \int_{0}^{t} x^{m}(j-x)^{n} dx = \int_{0}^{\frac{t}{j}} j^{m+n+1} x^{m}(1-x)^{n} dx = j^{m+n+1} \beta\left(\frac{t}{j}; m+1, n+1\right) $$

This readily yields the proof of (10). The proof of (7) also follows trivially once we set $t = 1$ in the last line of the equality.

$$ \text{THE INTEGRAL } \int_{0}^{t} x^{p} \cdot \beta\left(\frac{x}{j}; m, n\right) dx $$

Consider the integral

$$ I_2 = \int_{0}^{t} x^{p} \cdot \beta\left(\frac{x}{j}; m, n\right) dx $$
where $t \in [0, 1]$ and $j \in \mathbb{R} \setminus \{0\}$. We evaluate $I_2$ using integration by parts.
We have

$$I_2 = \int_0^t x^p \beta \left( \frac{x}{j}; m, n \right) dx = \beta \left( \frac{x}{j}; m, n \right) \left. \frac{x^{p+1}}{p+1} \right|_0^t - \frac{1}{j^{n+m-1}(p+1)} \int_0^t x^{m+p}(j-x)^{n-1} dx \\ \implies I_2 = \beta \left( \frac{t}{j}; m, n \right) \frac{t^{p+1}}{p+1} - \frac{j^{p+1}}{p+1} \beta \left( \frac{t}{j}; m+p+1, n \right)$$

This readily yields (11) and by setting $j = 1$, we have a proof of (9).

REFERENCES

[1] Necdet Batir and Anthony Sofo. Finite sums involving reciprocals of the binomial and central binomial coefficients and harmonic numbers. *Symmetry*, 13(11), 2021.

[2] Jacques Dutka. The incomplete beta function—a historical profile. *Archive for history of exact sciences*, pages 11–29, 1981.

[3] David Osborn and Richard Madey. The incomplete beta function and its ratio to the complete beta function. *Mathematics of Computation*, 22(101):159–162, 1968.

[4] B Sury, Tianming Wang, and Feng-Zhen Zhao. Identities involving reciprocals of binomial coefficients. *J. Integer Seq*, 7(2):04–2, 2004.

[5] Eric W Weisstein. Beta function. https://mathworld.wolfram.com/, 2002.

[6] Eric W Weisstein. Incomplete beta function. https://mathworld.wolfram.com/, 2003.
where $t \in [0, 1]$ and $j \in \mathbb{R} \setminus \{0\}$. We evaluate $I_2$ using integration by parts.
We have

$$I_2 = \int_0^t x^p \beta \left( \frac{x}{j}; m, n \right) dx = \beta \left( \frac{x}{j}; m, n \right) \left. \frac{x^{p+1}}{p+1} \right|_0^t - \frac{1}{j^{n+m-1}(p+1)} \int_0^t x^{m+p}(j-x)^{n-1} dx \\ \implies I_2 = \beta \left( \frac{t}{j}; m, n \right) \frac{t^{p+1}}{p+1} - \frac{j^{p+1}}{p+1} \beta \left( \frac{t}{j}; m+p+1, n \right)$$

This readily yields (11) and by setting $j = 1$, we have a proof of (9).

REFERENCES

[1] Necdet Batir and Anthony Sofo. Finite sums involving reciprocals of the binomial and central binomial coefficients and harmonic numbers. *Symmetry*, 13(11), 2021.

[2] Jacques Dutka. The incomplete beta function—a historical profile. *Archive for history of exact sciences*, pages 11–29, 1981.

[3] David Osborn and Richard Madey. The incomplete beta function and its ratio to the complete beta function. *Mathematics of Computation*, 22(101):159–162, 1968.

[4] B Sury, Tianming Wang, and Feng-Zhen Zhao. Identities involving reciprocals of binomial coefficients. *J. Integer Seq*, 7(2):04–2, 2004.

[5] Eric W Weisstein. Beta function. https://mathworld.wolfram.com/, 2002.

[6] Eric W Weisstein. Incomplete beta function. https://mathworld.wolfram.com/, 2003.