Coefficients involved in the Binomial Theorem. $\binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.

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Approximating the logarithm of the binomial coefficient

We know that by using Stirling approximation: $\log n! \approx n \log n$ So how to approximate $\log {m \choose n}$?
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Binomial Identity $\sum\binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$

I'm looking for a reference with the proof of the following binomial identity: $$\sum_{k=0}^n \binom{2n+1}{2k+1}\binom{m+k}{2n} = \binom{2m}{2n}$$ I've looked in a number of textbooks that have a ...
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Simplify $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$

Can this sum be simplified: $\sum \limits_{k=0}^{n} \binom{n}{k} 2^{\sqrt{k}}$ Or at least is there a simple fairly tight upperbound? EDIT So I think this sum is more easily bounded than I ...
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Binomial Theorem Identities

What's the actual difference between these two formulas (they're both in the chapter regarding binomial theorem). They're from two different textbooks : $${n\choose k}+{n\choose k+1}={n+1\choose ...
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Can this sum be simplified: $ \sum_{k=0}^{n-1} { n -1 \choose k } (-2)^{k} (2n - k)! $?

Can this expression be further simplified : $ \sum_{k=0}^{n-1} { n -1 \choose k } (-2)^{k} (2n - k)! $? This is the coefficient of $x^{2n}$ in the formal power series expansion of $(1-2x)^{n-1} \times ...
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How can I prove the identity $2(n-1)n^{n-2}=\sum_k\binom{n}{k}k^{k-1}(n-k)^{n-k-1}$?

How can I prove the identity $$2(n-1)n^{n-2}=\sum_k\binom{n}{k}k^{k-1}(n-k)^{n-k-1}?$$ I know that the number of trees on $n$ vertices is $n^{n-2}$, and that a tree with $n$ vertices has $n-1$ ...
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An upper bound for $\sum_{i = 1}^m \binom{i}{k}\frac{1}{2^i}$?

Does anyone know of a reasonable upper bound for the following: $$\sum_{i = 1}^m \frac{\binom{i}{k}}{2^i},$$ where we $k$ and $m$ are fixed positive integers, and we assume that $\binom{i}{k} = 0$ ...
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669 views

Sum of product of binomial coefficient

Is the following true? $$\sum_{x_1+x_2+...+x_n=n}\ \ \, \prod_{i=1}^{n}{k_i\choose x_i}={\sum_{i=1}^{n}k_i \choose n} .$$ I tried to use the multinomial theorem, but it doesn't seem applicable.
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Obtaining binomial coefficients without “counting subsets” argument

I want to obtain the formula for binomial coefficients in the following way: Elementary ring theory shows that $(X+1)^n\in\mathbb Z[X]$ is a degree $n$ polynomial, for all $n\geq0$, so we can write ...
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Proof of the identity $\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}$

Prove the identity: $$\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$
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Derive a closed form for a sum with inverse binomial coefficients

First off, I would like to apologize again for the integral I posted several days ago involving $\zeta(5)$. I was careless and did not examine the decimals out far enough. With that said, I would ...
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217 views

Bounding ${(2d-1)n-1\choose n-1}$

Claim: ${3n-1\choose n-1}\le 6.25^n$. Why? Can the proof be extended to obtain a bound on ${(2d-1)n-1\choose n-1}$, with the bound being $f(d)^n$ for some function $f$? (These numbers ...
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353 views

Problem of limit with binomial coefficients

I thought that the following would made a nice exercise, but I am not sure how to evaluate its difficulty since I often miss elementary solutions. How about you try answering it? It would be great to ...
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512 views

Closed form for a sum involving binomial coefficients

Let $n,k$ be positive integers. Is there a closed form of the sum $$\sum_{s=0}^{k} \binom{n}{s} \binom{s}{k-s}\text{?}$$ By that I mean a representation which is free of sums and hypergeometric ...
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An extrasensory perception strategy :-)

Inspired by classical Joseph Banks Rhine experiments demonstrating an extrasensory perception (see, for instance, the beginning of the respective chapter of Jeffrey Mishlove book “The Roots of ...
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Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$

Prove that $$ \frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n, \quad \forall k\in\mathbb{N}. $$ I tried already by induction over $k$ but i ...
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Consecutive terms in Pascal's Triangle

is it known whether or not there are infinitely many pairs of consecutive terms in this sequence: http://oeis.org/A006987 ? The sequence is the list of numbers expressible in the form ...
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134 views

Binomial formula for $(x+1)^{1/3}$ (related to Newton's binomial theorem)

I know that $$\displaystyle \sqrt{1+x} = \sum_{j=0}^{\infty}\left( \frac{(-1)^{(j-1)}}{2^{2j-1}\cdot(2j-1)}\binom{2j-1}{j}x^j\right). $$ Now, I want to evaluate $\sqrt[3]{1+x}$ but stuck at some ...
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Techniques for summing ratio of binomial coefficients

There are several identities that involve the sum of the product of binomial coefficients. However what I am searching for is an identity that involves the ratio of binomial coefficients. ...
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353 views

Evaluate a finite sum with four factorials

Given positive integers $k, m, n$ such that $1 \leq k \leq m \leq n$. Evaluate $$ \sum^{n}_{i\mathop{=}0}\frac{1}{n+k+i}\cdot\frac{(m+n+i)!}{i!(n-i)!(m+i)!}$$ Any hints? I'm stuck on ...
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Why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k} = 0$?

I know that the expansion of $\sum \limits_{k = 0}^{n} (-1)^{k} \binom{n}{k}$ equals to zero. But why is $\sum \limits_{k = 0}^{n} (-1)^{k} k\binom{n}{k}$ also equal to zero for $n \geq 2$? I've been ...
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Does $\sum_{k=0}^{k=n} {n \choose k} k!$ have a closed form for integers $k,n$?

While doing research in computer system, I came across the following summation: $$S_n = \sum_{k=0}^{n} {n \choose k} k! = \sum_{k=0}^{n} \frac{n!}{(n-k)!}$$ where both $n$ and $k$ are integers. $S_n$ ...
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Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$

I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$. I suppose I could use the counting rules in probability, perhaps combination= ...
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Intuitive explanation for a polynomial expansion?

Is there an ituitive explanation for the formula: $$ \frac{1}{\left(1-x\right)^{k+1}}=\sum_{n=0}^{\infty}\left(\begin{array}{c} n+k\\ n \end{array}\right)x^{n} $$ ? Taylor expansion around x=0 ...
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207 views

A peculiar binomial coefficient identity

While inventing exercises for a discrete math text I'm writing I came up with this $$ \binom{\binom{n}{2}}{2}=3\binom{n+1}{4} $$ It's an easy result to prove, but it got me wondering Is this pure ...
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Is there a combinatorial way to see the link between the beta and gamma functions?

The Wikipedia page on the beta function gives a simple formula for it in terms of the gamma function. Using that and the fact that $\Gamma(n+1)=n!$, I can prove the following formula: $$ ...
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prime numbers in Pascal's triangle

Just wondering about this: Is it true that there are no prime numbers in Pascal's triangle, with the exception of $\binom{n}{1}$ and $\binom{n}{n-1}$? From the first 18 lines it appears that this is ...
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Putting ${n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ in a closed form

As the title says, I'm trying to transform $\displaystyle{n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ into a closed form. My work: $\displaystyle\left(1 + ...
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Polynomial in $\mathbb{Q}[x]$ sending integers to integers?

We can view the binomial coefficient $\binom{x}{k}$ has a polynomial in $x$ with degree $k$. So taking some $f\in\mathbb{Q}[x]$, why is $f(n)\in\mathbb{Z}$ for all $n\in\mathbb{Z}$, precisely when the ...
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Non-probabilistic proofs of a binomial coefficient identity from a probability question

Combining the answers given by me and Ralth to the probability question at Probability Question, we get the following identity: $$ \sum\limits_{k = m}^n {{n \choose k}p^k (1 - p)^{n - k} {k \choose ...
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Show that ${-n \choose i} = (-1)^i{n+i-1 \choose i} $

Show that ${-n \choose i} = (-1)^i{n+i-1 \choose i} $. This is a homework exercise I have to make and I just cant get started on it. The problem lies with the $-n$. Using the definition I get: $${-n ...
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Binomial coefficients: how to prove an inequality on the $p$-adic valuation?

In section 4 of the article by Afred van der Poorten's A Proof That Euler Missed ... the following inequality is used: $$\nu_{p}\displaystyle\binom{n}{m}\leq\left\lfloor\dfrac{\ln n}{\ln ...
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Ordinary generating function for $\binom{3n}{n}$

The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}}$$ follows from the generalized ...
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Prove that $\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$

Prove that $$\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$$ What should I do for this equation? Should I focus on proving ...
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Combinatorial Proof of $\binom{\binom{n}{2}}{2} = 3 \binom{n}{3}+ 3 \binom{n}{4}$ for $n \geq 4$

For $n \geq 4$, show that $\binom{\binom{n}{2}}{2} = 3 \binom{n}{3}+ 3 \binom{n}{4}$. LHS: So we have a set of $\binom{n}{2}$ elements, and we are choosing a $2$ element subset. RHS: We are ...
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Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$

This result, $$\prod_{k=1}^{\infty} \big\{\big(1+\frac1{k}\big)^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$$ is in a paper by Hirschhorn in the current issue of the Fibonacci Quarterly (vol. ...
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Asking About Binomial Sum Related to Fibonacci

How would I prove $$ \sum\limits_{i,j\ge 0} {n-i \choose j} {n-j \choose i}=F_{2n+1} $$ where $n$ is a nonnegative integer and $\{F_n\}_{n\ge 0}$ is a sequence of Fibonacci numbers? Thank you very ...
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280 views

Smallest constant in exponent so that limit of sum is $0$

I am trying to work out the smallest constant $c>0$ so that $$\lim_{n \to \infty} \sum_{a=1}^n \sum_{b=0}^n {n \choose a} {n-a \choose b} \left({a+b \choose a} 2^{-a-b}\right)^{c n/\ln{n}} =0 .$$ ...
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How can we find the gcd for elements (binomial coefficient)?

$\gcd\left(\binom{2n}1,\binom{2n}3,\binom{2n}5,\ldots,\binom{2n}{2n-1}\right)$ i want to know what is specialty of such a series.I am not able to generalize the problem solution.Is there any rule for ...
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Summation of an Infinite Series: $\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$

I am having trouble proving that $$\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$$ I know that $$\frac{2x \ \arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=1}^\infty ...
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A three variable binomial coefficient identity

I found the following problem while working through Richard Stanley's Bijective Proof Problems (Page 5, Problem 16). It asks for a combinatorial proof of the following: $$ \sum_{i+j+k=n} ...
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Binomial identity

I'd like to get a hint to prove the following identity: $$\tag{1}\sum_{\nu}(-1)^{\nu}\displaystyle \binom{a}{\nu}\binom{n-\nu}{r}=\binom{n-a}{n-r} .$$ The original statement reads "By specialization ...
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Sum of squares of binomial coefficients

I came across the following sum in reference to this question $$\sum_{n=0}^{\infty} \frac{1}{2^{5 n}} \binom{2 n}{n}^2 = \frac{\sqrt{\pi}}{\Gamma \left( \frac{3}{4}\right)^2}$$ The sum on the left ...
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Computing the last non-zero digit of ${1027 \choose 41}$?

I am working on the following problem: Let $x_n$ be a sequence of positive odd numbers. If $N$ is the number of ordered pairs $(x_1, x_2, x_3, \dots, x_{42})$ such that $$x_1 + x_2 + x_3 + \dots + ...
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184 views

A sum with binomial coefficients

Show that $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^{n}n(n+2)!}{6}.$$
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1answer
188 views

Evaluate $\lim_{n \to \infty} \sum_{j=0}^{n}{{j+n-1} \choose j}\frac{1}{2^{j+n}}$

Evaluate $$\lim_{n \to \infty} \sum_{j=0}^{n}{{j+n-1} \choose j}\frac{1}{2^{j+n}}$$ I don't understand where to start. Please help.
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337 views

Derivation of bound on expression involving binomial coefficient from Erdős and Rényi 1959

I'm in the process of working through Erdős and Rényi's 1959 article "On Random Graphs I". In the proof of the first Lemma, equation 14 gives a bound on an expression involving several binomial ...
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1answer
422 views

$\binom{n}{k} : \binom{n}{k+1} : \binom{n}{k+2} = a : b : c$

It is a rather surprising fact (to me, at least) that $\displaystyle \binom{14}{4} = 1001$; $\displaystyle \binom{14}{5} = 2002$; $\displaystyle \binom{14}{6} = 3003$. Actually, this is the only ...
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How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$?

I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am ...
7
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3answers
133 views

Inequality $\binom{2n}{n}\leq 4^n$

I would like to prove the following inequality, for $n=0,1,2,...$, $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.