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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1answer
695 views

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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6answers
599 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
261 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.
10
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4answers
202 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.
10
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3answers
406 views

Evaluating a sum involving binomial coefficient in denominator

I came across the following sum: $$\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2}\frac{4^k}{{2k \choose k}}$$ I thought that this can be evaluated using the expansion of ...
10
votes
3answers
388 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 ? $$\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \binom{n}{k} = \sum_{k=1}^{n} \frac{1}{k}$$
10
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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 ?
10
votes
3answers
245 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..?
10
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1answer
438 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
342 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 ...
10
votes
1answer
786 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.
10
votes
6answers
190 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.
10
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2answers
128 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.
10
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3answers
404 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 ...
10
votes
1answer
224 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
votes
3answers
614 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 ...
10
votes
1answer
149 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 ...
9
votes
6answers
322 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
445 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 ...
9
votes
3answers
538 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 ...
9
votes
3answers
826 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
446 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 ...
9
votes
2answers
439 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
2answers
337 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
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3answers
528 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
288 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: $$\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$
9
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4answers
241 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
3answers
135 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. ...
9
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2answers
466 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
2answers
385 views

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 ...
9
votes
1answer
167 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 ...
9
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2answers
188 views

An asymptotic expression of sum of powers of binomial coefficients.

Let $k$ be a fixed positive number and $n$ an integer increasing to infinity. Then $$\sum_{\nu =0}^n \binom{n}{\nu}^k \sim \frac{2^{kn}}{\sqrt{k}} \left( \frac{2}{\pi n} \right)^{\frac{k-1}{2}}.$$ ...
9
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2answers
249 views

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 ...
9
votes
4answers
300 views

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 ...
9
votes
2answers
192 views

Number of ways to arrange $n$ items in $m$ positions having exactly $k$ items adjacent to each other

It was over 20 years since I studied maths and I am stuck. I'd really appreciate some help understanding this (probably quite simple) problem. I have $n$ items that I can place on $m$ positions. $m$ ...
9
votes
1answer
102 views

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 ...
9
votes
1answer
145 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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2answers
429 views

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. ...
9
votes
1answer
357 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
votes
0answers
133 views

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
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4answers
388 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 ...
8
votes
4answers
373 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
252 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
votes
7answers
859 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
196 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
324 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
3answers
334 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
500 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
866 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 ...