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

$\sum_{k=0}^{n} \binom{2n}{k}^2$ So i'm trying to simplify this one and I'm stuck in nowhere. Some kind of tip would be appreciated. Thanks! :)
11
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3answers
216 views

Calculating $\sum_{0\le k\le n/2} \binom{n-k}{k}$

I would like to evaluate: $$\sum_{0\le k\le n/2}\binom{n-k}{k}$$ Any idea?
11
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4answers
439 views

Show that if $\prod\limits_{k=1}^{n}(x+a_k)=\sum\limits_{k=0}^{n} {n\choose k}a^k_kx^{n-k}$ then $a_1=a_2=a_3=…=a_{n-1}=a_n$

Let $a_0=1$. Prove that, if $$\prod_{k=1}^{n}(x+a_k)=\sum_{k=0}^{n} {n\choose k}a^k_kx^{n-k}=x^n+{n\choose 1}a_1x^{n-1}+{n\choose 2}a^2_2x^{n-2}+....+a^n_n,$$ then ...
11
votes
4answers
300 views

Determine $\displaystyle \lim_{n \to \infty}{{n} \choose {\frac{n}{2}}}\frac{1}{2^n}$, where each $n$ is even

For each positive even integer $n$, set $$P_n = \displaystyle {{n} \choose {\frac{n}{2}}}\frac{1}{2^n}.$$ Show that $\displaystyle \lim_{n \to \infty} P_n$ exists and determine its value. Here's ...
11
votes
4answers
903 views

Combinatorial proof that binomial coefficients are given by alternating sums of squares?

A student recently asked whether there was a combinatorial proof of the following identity: $\begin{equation*} \sum^n_{k=1}(-1)^{n-k}k^2 = {n+1 \choose 2}. \end{equation*}$ I was in a rush and ...
11
votes
5answers
240 views

Sum $\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$

$\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}=\sum_{n=i}^{\infty} \frac {1}{{2n \choose n-i}}$ is a very interesting one. Here is what I have from WolframAlpha. $\displaystyle ...
11
votes
3answers
199 views

Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$

$$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$ Class themes are: Generating functions and formal power series.
11
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2answers
282 views

The Fermat prime 257 and binomial sum $\sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}}$?

We have, $\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{27}(9-\pi\sqrt{3}\,)\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - ...
11
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2answers
559 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 ...
11
votes
3answers
631 views

Prove the lecturer is a liar…

I was given this puzzle: At the end of the seminar, the lecturer waited outside to greet the attendees. The first three seen leaving were all women. The lecturer noted " assuming the attendees are ...
11
votes
1answer
126 views

How find all postive integer number such $(n+k)\nmid \binom{2n}{n}$

Question: Find the all integer $k$,such there are exist infinitely many $n$ such $$(n+k)\nmid \binom{2n}{n}$$ This is china 2014 (CMO problem 4),it's have been end exam three hours ago. I ...
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3answers
315 views

Closed form for $\sum_{k=0}^{n} k\binom{n}{k}\log\binom{n}{k}$

Is it possible to write this in closed form: $$\sum_{k=0}^{n} k\binom{n}{k}\log\left(\vphantom{\Huge A}\binom{n}{k}\right)$$ Can you get something like $$n2^{n-1}\log(2^{n-1})$$
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1answer
122 views

Asymptotic Behavior of a Sum with Binomial Coefficients

The Problem: Find the asymptotic behavior (with respect to $n$) of the following sum $$\sum\limits_{j = 3}^n \binom{n}{j} \frac{(j - 1)!}{2\cdot n^j}. $$ Where the Problem Comes From: If we ...
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1answer
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Factorial canceling on expansion of binomial coefficients on Concrete Mathematics

On Concrete Mathematics section 5.5, which is teaching the hypergeometric functions, generalized factorials is defined as: \[ \frac 1 {z!} = \lim_{n \to \infty} \binom{n+z}{n}n^{-z} \] where \[ ...
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votes
6answers
608 views

Evaluate $ \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots$ [duplicate]

I need to evaluate, for a certain worded question: If n is even $$\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n}$$ If n is odd ...
10
votes
6answers
272 views

Why, conceptually, is it that $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$? [duplicate]

Why, conceptually, is it that $$\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}?$$ I know how to prove that this is true, but I don't understand conceptually why it makes sense.
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4answers
216 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.
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3answers
404 views

Help with combinatorial proof of identity: $\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^{n} \frac{1}{k}$

How to prove this identity? Can someone please give me some insight ? $$\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^{n} \frac{1}{k}$$
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3answers
625 views

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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2answers
1k views

Good upper bound for $\sum\limits_{i=1}^{k}{n \choose i}$?

I want an upper bound on $$\sum_{i=1}^k \binom{n}{i}.$$ $O(n^k)$ seems to be an overkill -- could you suggest a tighter bound ?
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3answers
260 views

How to find sums like $\sum_{k=0}^{39} \binom{200}{5k}$

How do I find sums like these?-- $$S=\displaystyle\sum_{k=0}^{39} \dbinom{200}{5k}$$ that is, when there is a summation of binomial coefficients, but with jumps of some terms..?
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1answer
447 views

New Year Combinatorics

In the spirit of the festive period and in appreciation of the encouraging response to my X'mas Combinatorics problem posted recently, here's one for the New Year! Express the following as a ...
10
votes
2answers
345 views

The integral $\int_0^1 \frac{(x+1)^n-1}{x} dx$

I know that the integral $\int_0^1 \frac{(x+1)^n-1}{x} dx,$ for $n \in \mathbb{Z}^+$, can be evaluated by expanding the numerator with the binomial theorem and integrating term by term. You get the ...
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1answer
830 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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324 views

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: $$\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$
10
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6answers
194 views

A limit on binomial coefficients

Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$. What I can do is just use Stolz formula. But I could not proceed.
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3answers
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proof that $1 = \sum\limits_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k}$

I'm looking for a proof of this identity: $$ 1 = \sum_{k=0}^n (-1)^k { 2n \choose n,k,n-k } \frac{n}{n+k} $$ I'll take anything, but a combinatorial proof would be nice - all of the terms in the sum ...
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2answers
132 views

Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $\sum_{k=1}^n\frac{a_k-b_k}k$

Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $$\sum_{k=1}^n\frac{a_k-b_k}k$$ By computing some partial sums, the answers are 0. It seems an inductive argument is possible.
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3answers
417 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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1answer
226 views

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

Can someone please help me simplifying this sum $$\sum_{k=0}^n \binom{n}{k} 2^{k^2}$$ Wolframalpha fails (see here). Thanks in advance. The sum counts the number of (labelled) digraphs (with ...
10
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1answer
180 views

Prove $\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$ for-

Let $n$ be a positve integer. Prove that$$\sum\limits^m_{k=0} \frac{2n-k\choose k}{2n-k\choose n}\frac{2n-4k+1}{2n-2k+1}2^{n-2k}=\frac{n\choose m}{2n-2m\choose n-m}2^{n-2m}$$ for each non-negative ...
10
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1answer
176 views

Interpretation of a combinatorial identity

I am trying to find an combinatorial interpretation for the following combinatorial identity involving iterated binomial coefficients, which appeared in the November 1980 edition of The American Math ...
10
votes
1answer
153 views

Can this product be written so that symmetry is manifest?

Let $i,$ $j,$ $k$ be nonnegative integers such that $i+j+k$ is even. The expression $$(-1)^{j+k}\binom{i+j+k}{i,j,k}\prod_{\ell=0}^{k-1} \frac{i-j+k-2\ell-1}{i+j+k-2\ell-1}$$ apparently computes the ...
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2answers
229 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 ...
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0answers
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Is the given binomial sum almost everywhere negative as $K\to\infty$?

The binomial sum is as follows: $$\mathcal {L}^K(\theta)= \sum_{i=\lceil{K/2}\rceil}^K \binom{K}{i}\theta^i\left((1-\theta)^{K-i}-\frac{1}{2}(1-\theta)^{-K}(1-2\theta)^{K-i}\right)$$ which can also ...
9
votes
6answers
346 views

Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $

Help me to simplify:$$\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $$ I got a hunch that it will depend on whether $n$ is a multiple of $6$ and equals to $\frac{2^n+2}{3}$ when $n$ is a ...
9
votes
5answers
483 views

factorial as difference of powers: $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$?

The successive difference of powers of integers leads to factorial of that power. I think this is the formula $\sum_{r=0}^{n}\binom{n}{r}(-1)^r(l-r)^n=n!$ But I found no proof on internet. Please ...
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4answers
340 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 ...
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3answers
831 views

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 ...
9
votes
4answers
482 views

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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votes
2answers
453 views

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 ...
9
votes
1answer
525 views

How to compute the asymptotic growth of $\binom{n}{\log n}$?

I'm interested with tight bounds for: $$f(n)={n\choose{\log{n}}}$$ It sounds like it's something simple, but I can't get a nice expression I can use. Any ideas on how to do this?
9
votes
2answers
339 views

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 $n-1$ ...
9
votes
4answers
241 views

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 $\binom{i}{k} = 0$ ...
9
votes
4answers
276 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 ...
9
votes
3answers
609 views

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 ...
9
votes
2answers
477 views

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 ...
9
votes
4answers
243 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 ...
9
votes
2answers
78 views

When is a binomial coefficient a factorial, i.e. when is $\binom{m}{j} = n!$ for some $n,m,j$?

As stated in the title: when is a binomial coefficient a factorial, i.e. when is $\binom{m}{j} = n!$ for some $m,j,n$? I was thinking about this problem a couple of days ago because in all my years of ...
9
votes
3answers
136 views

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. ...