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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Showing that $\lceil (\sqrt{3} + 1)^{2n} \rceil$ is divisible by $2^{n+1}$.

I have a question which has fluxommed me and my pals for the past few days. Any help or solution is welcome Show using Binomial theorem that the integer just after $(3^{1/2} + 1)^{2n}$ is divisble ...
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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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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 ...
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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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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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The binomial formula and the value of 0^0

Here is the text from Knuth's The Art of computer programming, 1.2.6 F formula 14: Knuth doesn't give the proof of the statement. So, I tried to write it myself. To make binomial formula equal to ...
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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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Combinatorics question: Show divisibility

Let $a\geq2$, $b\geq2$ be two prime numbers and k be a natural number with $k\leq min(a,b)$. How can one show that $z := \binom{a+b}{k} - \binom{a}{k} - \binom{b}{k}$ is divisible by the product ...
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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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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?
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Squared binomial coefficient

I've got the following finite sum: $s_{n}=\sum\limits_{k=0}^{n}\binom{n}{k}^2p^k$ (esp. if $p$ is a function of $n$, like $p=\frac1{n}$), which can be rewritten as ...
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247 views

Expansion of $ (a_1 + a_2 + \cdots + a_k)^n $

Is there an expansion for the following summation? $$ (a_1 + a_2 + \cdots + a_k)^n $$
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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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Combinatorial argument for the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$

I am looking for the combinatorial argument for the identity: \begin{equation} k\binom{n}{k} = n\binom{n-1}{k-1} \end{equation} This is easy to show algebraically as: \begin{equation} \binom{n}{k} ...
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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} ...
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How to count matrices with rows and columns with an odd number of ones?

I proved that $\displaystyle \left(\sum_{k\, \rm odd}\binom{m}{k}\right)^{n-1}=\left(\sum_{k\;{\rm odd}}\binom{n}{k}\right)^{m-1}$ by counting matrices of size $n\times m$ with entries in $\{0,1\}$ ...
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Simplify $\sum_{k=1}^{n} {k\choose m} {k}$

$\sum_{k=1}^{n} {k\choose m} {k}$ I have tried to expand it, but the m is pretty annoying. Any ideas to get rid of the summation and give a simple formula? There is a part before $\sum_{k=1}^{n} ...
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Closed-form expression for $\sum_{n=1}^{k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}$?

Wolframalpha tells me that $$\sum_{n=1}^{k} (-1)^{n+1}n^2(n^2-1)\binom{2k}{k-n}=0$$ for $k>2$ However I have not been able to come up with a proof and I simply don't see how to do it. Does anyone ...
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Combinatorial Identity $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$

Show that $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$. In the LHS $\binom{n+r-1}{r}$ counts the number of ways of selecting $r$ objects from a set of size $n$ where ...
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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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Cousin of the Vandermonde binomial identity

The Vandermonde binomial identity can be expressed as \begin{align*} \sum_{i+j=r} \binom{m}{i} \binom{n}{j} = \binom{m+n}{r} && r \leq m +n. \end{align*} While working on an algebra problem, I ...
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262 views

Bounds on $\sum_{k=0}^{m} \binom{n}{k}x^k$ and $\sum_{k=0}^{m} \binom{n}{k}x^k(1-x)^{n-k}, m<n$

I've read this interesting article by Woersch (1994) dealing with approximation of binomial coefficients (rows of Pascal's triangle). I'm just wondering if similar bounds exist for partial binomial ...
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301 views

A binomial coefficient identity?

Suppose $p$, $k$ and $s$ are integers with $s,k \le p$. Consider the following polynomial in $x$ and $y$, $$ \sum_{\ell=0}^k \binom{s}{\ell} \binom{p-s}{k-\ell} x^\ell y^{p-\ell}$$ Does this ...
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A finite sum involving the binomial coefficients and the harmonic numbers

Wikipedia has a proof of the identity $$ H_{n} =\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k}$$ http://en.wikipedia.org/wiki/Harmonic_number#Calculation Curiously, there is also the identity ...
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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 ...
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Determining when a certain binomial sum vanishes

Consider the following sum of signed binomial coefficients: $$S_{n,a,p} = \sum_{i \equiv a \mod p} \binom{n}{i}(-1)^i$$ ($n$ is a positive integer, $p$ is an odd prime, $a$ is between $0$ and ...
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Asymptotic difference between a function and its binomial average

The origin of this question is the identity $$\sum_{k=0}^n \binom{n}{k} H_k = 2^n \left(H_n - \sum_{k=1}^n \frac{1}{k 2^k}\right),$$ where $H_n$ is the $n$th harmonic number. Dividing by $2^n$, we ...
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Does this really converge to 1/e? (Massaging a sum)

Short version: can we prove that $$\sum_{k=0}^n (-1)^k \binom{n}{k}^2 \frac{k!}{n^{2k}} \to \frac1e$$ as $n \to \infty$? Long version: First, consider $$a_n = \sum_{k=0}^n \frac{(-1)^k}{k!}$$ It is ...
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asymptotics of the expected number of edges of a random acyclic digraph with indegree and outdegree at most one

A recent discussion, which may be found here, examined the problem of counting the number of acyclic digraphs on $n$ labelled nodes and having $k$ edges and indegree and outdegree at most one. It was ...
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Combinatorial Interpretation of Fractional Binomial Coefficients

My question is a bit imprecise - but I hope you like it. I even strongly think it has a proper answer. The binomial coefficient $\binom{\frac{1}{2}}{n}$ is strongly related to Catalan numbers - the ...
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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 ...
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Asymptotics for sum of binomial coefficients from Concrete Mathematics

Concrete Mathematics EXERCISE 9.25: Supposing \[ S_n = \sum_{k=0}^n \binom{3n}k \] Prove that \[ S_n = \binom{3n}{n}\left(2-\frac4n+O\left(\frac1{n^2}\right)\right) \] This sequence also ...
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Find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$?

I just want to ask you if my solution is correct. Here's the problem, Using the Binomial Theorem, find the coefficient of $\sqrt{3}$ in $(1+\sqrt{3})^7$. Solution: The binomial theorem is, ...
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Simplify $\sum_{i=0}^n (i+1)\binom ni$

Simplifying this expression$$1\cdot\binom{n}{0}+ 2\cdot\binom{n}{1}+3\cdot\binom{n}{2}+ \cdots+(n+1)\cdot\binom{n}{n}= ?$$ $$\text{Hint: } \binom{n}{k}= \frac{n}{k}\cdot\binom{n-1}{k-1} $$
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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 $n,\,p$ ...
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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 ...
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Find the sum of this series :$ \frac{1}{{1!2009!}} + \frac{1}{{3!2007!}} + \cdots + \frac{1}{{1!2009!}}$

Find the sum of this series : $$\sum\limits_{\scriptstyle 1 \leqslant x \leqslant 2009 \atop {\scriptstyle x+y=2010 \atop \scriptstyle {\text{ }}x,y{\text{ odd}} }} {\frac{1}{{x!y!}}} = ...
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The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is

The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is? I tried this: ...
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Short and intuitive proof that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$

The simple inequality that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$ has a number of different proofs. But is there a particularly intuitive, short and elegant proof that uses the natural ...
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Is there a binomial identity for this expression $\frac{\binom{r}{k}}{\binom{n}{k}}$?

What I'm trying to prove is this summation: $$\sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i = \dfrac{r}{n} \cdot k$$ I used induction on $k$ as follows: ...
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Number of acyclic digraphs on $[n]$ with $k$ edges and each indegree, outdegree $\leq 1$

How many (labelled) acyclic digraphs are there on the vertex set $[n]$ with exactly $k$ edges and each indegree, outdegree $\leq 1$? The answer is $${n \choose k} {n-1 \choose k} k!.$$ Is there a ...
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How to understand this combinatorially: $\sum^{2k}_{i=0} \binom{4k}{2i} (-1)^{i}=2^{2k}(-1)^{k}$

The TAs in my department are stuck in assisting an undergraduate with the following problem: $$\sum^{2k}_{i=0} C^{4k}_{2i}(-1)^{i}=2^{2k}(-1)^{k}.$$ We tried to solve this via induction (obviously ...
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Why does this expected value simplify as shown?

I was reading about the german tank problem and they say that in a sample of size $k$, from a population of integers from $1,\ldots,N$ the probability that the sample maximum equals $m$ is: ...
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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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Bounds for $\binom{n}{cn}$ with $0 < c < 1$.

Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$? I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn} \leq ...
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Inductive proof for the Binomial Theorem for rising factorials

I want to proove the following equality containing rising factorials $$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$ For $n=1$ this equality is ...
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What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$?

What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$ in terms of $d$? Does the limit exist? Is there a simple upper bound interms of $d$?
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272 views

Inequality involving factorial

I am trying to prove following inequality: $$\binom{n}{k}<(en/k)^k$$ I tried Stirling approximation but I could not get anything further. Then I get $$\binom{n}{k}\approx \frac{\sqrt{2\pi ...
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183 views

Help evaluating a limit

I have the following limit: $$\lim_{n\rightarrow\infty}e^{-\alpha\sqrt{n}}\sum_{k=0}^{n-1}2^{-n-k} {{n-1+k}\choose k}\sum_{m=0}^{n-1-k}\frac{(\alpha\sqrt{n})^m}{m!}$$ where $\alpha>0$. ...
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696 views

partial sum involving factorials

Here is an interesting series I ran across. It is a binomial-type identity. $\displaystyle \sum_{k=0}^{n}\frac{(2n-k)!\cdot 2^{k}}{(n-k)!}=4^{n}\cdot n!$ I tried all sorts of playing around, ...