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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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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1answer
362 views

Construction of generating function from identity

I am trying to solve identity involving binomials and Fibonacci numbers by using generating functions: $$\sum_{k=0}^n{n \choose k}{n+k\choose k}f_{k+1}=\sum_{k=0}^n{n \choose k}{n+k\choose ...
9
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4answers
400 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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10answers
3k views

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 ...
8
votes
4answers
375 views

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 ...
8
votes
4answers
264 views

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. $S_n$ ...
8
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7answers
901 views

Prove the following equality: $\sum_{k=0}^n\binom {n-k }{k} = F_n$ [duplicate]

I need to prove that there is the following equality: $$ \sum\limits_{k=0}^n {n-k \choose k} = F_{n} $$ where $F_{n}$ is a n-th Fibonacci number. The problem seems easy but I can't find the way to ...
8
votes
4answers
203 views

Proving this binomial identity $\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$

A teacher gave this as a homework question, and I have tried but haven't been able to arrive at a solution. $\sum_{k=0}^n {n+k \choose k} \frac{1}{2^{k}}= 2^{n}$ Could someone prove it, or at least ...
8
votes
3answers
325 views

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 ...
8
votes
2answers
94 views

Rough bound for sum $\binom{3n}{0}+\binom{3n}{1}+\cdots+\binom{3n}{n-1}$

Is it true that $$\frac{\dbinom{3n}{0}+\dbinom{3n}{1}+\cdots+\dbinom{3n}{n-1}}{2^{3n}}<\frac13$$ for all positive integers $n$? I've plotted the first few values of $n$ and noticed that the ...
8
votes
3answers
340 views

How to evaluate $\sum\limits_{k=0}^{n} \sqrt{\binom{n}{k}} $

Can we find $$ \sum_{k=0}^{n} \sqrt{\binom{n}{k}} \quad$$ This problem asked me my friend about a year ago, but I didn't know how to attack problem. Now, I am interesting in solution. Any suggestion? ...
8
votes
4answers
516 views

Show that $\sum_{k=0}^n\binom{3n}{3k}=\frac{8^n+2(-1)^n}{3}$

The other day a friend of mine showed me this sum: $\sum_{k=0}^n\binom{3n}{3k}$. To find the explicit formula I plugged it into mathematica and got $\frac{8^n+2(-1)^n}{3}$. I am curious as to how one ...
8
votes
3answers
900 views

Combinatorial proof for two identities [duplicate]

Does exist a combinatorial proof for the following two identities ? $\sum_{k = 0}^{n} \binom{x+k}{k} = \binom{x+n+1}{n}$ $\sum_{k = 0}^{n} k\binom{n}{k} = n2^{n-1}$ I know how to derive the ...
8
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3answers
233 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 ...
8
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2answers
498 views

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: $$ ...
8
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2answers
1k views

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 ...
8
votes
3answers
234 views

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 + ...
8
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3answers
379 views

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 ...
8
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4answers
480 views

Inequality with central binomial coefficients

For every even positive number $N$ we have: $$ {2N \choose N } < 2^N {N \choose N/2 } < 2 {2N \choose N } $$ (Furthermore, $\frac{2^N {N \choose N/2 }}{{2N \choose N }} \to \sqrt{2} $ for ...
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5answers
841 views

Proof of $\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1}$?

How do I prove that $$\sum_{0 \le k \le t} {t-k \choose r}{k \choose s}={t+1 \choose r+s+1} \>?$$ I saw this in a book discussing generating functions.
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3answers
410 views

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 ...
8
votes
4answers
318 views

Orthogonality for Binomial Coefficients

Could somebody explain to me where these two formulas come from as applications of the binomial theorem? $$\sum_{k=0}^n {n \choose k}(-1)^kk^r=0$$ for non-negative integers $r\lt n$. And ...
8
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3answers
102 views

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 ...
8
votes
2answers
733 views

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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3answers
392 views

A Curious Binomial Sum Identity without Calculus of Finite Differences

Let $f$ be a polynomial of degree $m$ in $t$. The following curious identity holds for $n \geq m$, \begin{align} \binom{t}{n+1} \sum_{j = 0}^{n} (-1)^{j} \binom{n}{j} \frac{f(j)}{t - j} = (-1)^{n} ...
8
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3answers
394 views

Proving the sum of squares of sine and cosine using the Cauchy product formula

Here are the power series of sine and cosine: $$\sin(x) = \sum_{n=0}^{\infty} (-1)^n \frac {x^{2n+1}} {(2n+1)!}$$ and $$\cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac {x^{2n}} {(2n)!}$$ How can it be ...
8
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2answers
216 views

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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4answers
267 views

Prove that $\sum_{k=0}^{m}\binom{m}{k}\binom{n+k}{m}=\sum_{k=0}^{m}\binom{n}{k}\binom{m}{k}2^k$ [duplicate]

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 ...
8
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1answer
305 views

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 ...
8
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2answers
131 views

An interesting property of binomial coefficients that I couldn't prove

So when I was trying to prove the argument in this link I've come up with something. When you extract the left term from the right term, you get the term under them. What is interesting is that as ...
8
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1answer
428 views

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 ...
8
votes
1answer
285 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 $$\lim_{n \to \infty} \sum_{a=1}^n \sum_{b=0}^n {n \choose a} {n-a \choose b} \left({a+b \choose a} 2^{-a-b}\right)^{c n/\ln{n}} =0 .$$ ...
8
votes
3answers
183 views

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 ...
8
votes
2answers
284 views

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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3answers
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Proving $\sum_{k=0}^n{2k\choose k}{2n-2k\choose n-k}=4^n$ [duplicate]

Some background. I was asked to find an arithmetic function $f$ such that $f*f=\mathbf 1$ where $\mathbf 1$ is the constant function 1 and $*$ denotes Dirichlet convolution. I was able to prove that ...
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2answers
296 views

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 + ...
8
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4answers
146 views

A binomial identity from Mathematical Reflections

Here is the problem: Let $m,n$ be positive integers with $n>m$. Prove that $\displaystyle\sum_{k=0}^{\lfloor\frac{n+m}2\rfloor} (-1)^{k}\binom{n}{k}\binom{m+n-2k}{n-1}=\binom{n}{m+1}$ This ...
8
votes
1answer
194 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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1answer
2k views

Asymptotics of binomial coefficients and the entropy function

I found a question while I was trying to practice Combinatorics and Probabilistic methods.I tried to solve it with no success.. this is the question: Use the Stirling approximation of the ...
8
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2answers
216 views

How prove binomial cofficients $\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$

How prove this $$\sum_{k=0}^{[\frac{n}{3}]}(-1)^k\binom{n+1}{k}\binom{2n-3k}{n}=\sum_{k=[\frac{n}{2}]}^n\binom{n+1}{k}\binom{k}{n-k}$$ This equation How prove it? Thank you I want take this ...
8
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1answer
302 views

Vandermonde identity in a ring

Let $R$ be a commutative $\mathbb{Q}$-algebra. For $r \in R$ and $n \in \mathbb{N}$ we can define the binomial coefficient $\binom{r}{n}$ as usual by $\binom{r}{0}=1$ and ...
8
votes
2answers
4k views

Inductive proof that ${2n\choose n}=\sum{n\choose i}^2.$

I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$ I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively ...
8
votes
2answers
187 views

$\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ implies $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2}$; where $p>3$ is a prime?

From $\binom{2p-1}{p-1}\equiv 1\pmod{\! p^2}$ how does one get $\binom{ap}{bp}\equiv\binom{a}{b}\pmod{\! p^2},\,\forall a,b \in \mathbb N,\, a>b$; where $p>3$ is a prime ?
8
votes
1answer
360 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 ...
8
votes
1answer
429 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 ...
8
votes
0answers
74 views

Generalizing Bellard's “exotic” formula for $\pi$ to $m=11$

Bellard's "exotic" pi formula has the form, $$a\pi+b = \sum_{n=1}^\infty \dfrac{P(n)}{{\displaystyle \binom{mn}{2n}2^{n-1}}}$$ where $a,b,m$ are integers and he uses $m=7$. However, it seems there ...
8
votes
0answers
160 views

Solution of $\large\binom{x}{n}+\binom{y}{n}=\binom{z}{n}$ with $n\geq 3$

I found this question in an old problem set. There's no hint or solution mentioned. For $n \geq 3$, prove or disprove the existence of $(x,y,z) \in \mathbb N^3, ...
7
votes
6answers
467 views

Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$

Is there a closed-form expression for the sum $\sum_{k=0}^n\binom{n}kk^p$ given positive integers $n,\,p$? Earlier I thought of this series but failed to figure out a closed-form expression in $n,\,p$ ...
7
votes
2answers
730 views

Finding $\lim\limits_{n \to \infty} \sum\limits_{k=0}^n { n \choose k}^{-1}$

We know that $$ 2^n= (1+1)^n = \sum_{k=0}^n {n \choose k}$$ I was asked to solve this limit, $$\lim_{n \to \infty} \ \sum_{k=0}^n {n \choose k}^{-1}=? \quad \text{for} \ n \geq 1$$
7
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5answers
303 views

The $n^{th}$ root of the geometric mean of binomial coefficients.

$\{{C_k^n}\}_{k=0}^n$ are binomial coefficients. $G_n$ is their geometrical mean. Prove $$\lim\limits_{n\to\infty}{G_n}^{1/n}=\sqrt{e}$$