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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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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2answers
346 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 ...
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257 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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2answers
188 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
191 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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166 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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1answer
342 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
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1answer
427 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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1answer
335 views

Construction of generating function from identity

I am trying to solve identity involving binomials and fibbonaci 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 ...
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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 ...
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719 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 ...
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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, $$\lim_{n \to \infty} \ \sum_{k=0}^n {n \choose k}^{-1}=? \quad \text{for} \ n \geq 1$$
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5answers
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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 $$\lim\limits_{n\to\infty}{G_n}^{1/n}=\sqrt{e}$$
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Prove that $\binom n2 + \binom {n-1}2$ is always a perfect square

Prove that if $n$ is a positive integer and $n >1$: $$\binom n2 + \binom {n-1}2$$ is always a perfect square. I know we need to turn that into a binomial, but I can't follow how. Please note I'm ...
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254 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 ...
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3answers
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Proof of a combinatorial identity: $\sum_{i=0}^n {2i \choose i}{2(n-i)\choose n-i} = 4^n$ [duplicate]

Possible Duplicate: Identity involving binomial coefficients This was part of a homework assignment that I had, and I couldn't figure it out. Now it is bugging me. Can anyone help me? ...
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Hard elementary combinatorics problem

How does one compute (without brute force) the smallest integer $n$ such that $\binom{2n}{1}(-3)^0 + \binom{2n}{3}(-3)^1 + \binom{2n}{5}(-3)^2 + \cdots + \binom{2n}{2n-1}(-3)^{(n-1)} = 0$?
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5answers
349 views

Are there surprisingly identical binomial coefficients?

Suppose $\binom{n}{k}=\binom{n'}{k'}$ with $k \geq 2$, $k' \geq 2$, $n \geq 2k$ and $n' \geq 2k'$. Does it follow that $n=n'$ and $k=k'$? EDIT: Yup, ...
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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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457 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 ...
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673 views

Summation Identity: $\sum_{i=1}^ni^3 = \left( \frac{n(n+1)}{2} \right)^2$

I have to prove: $$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2$$ Using the following: $$n^3 = 6 {n \choose 3} + 6 {n \choose 2} + n \quad \forall n \in \mathbb{N}$$ My work is that ...
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2answers
341 views

Asymptotics of $\sum_{k=0}^{n} {\binom n k}^a$

I need to estimate the asymptotics of $$\sum_{k=0}^{n} {\binom n k}^a, \quad a>2, \quad a \in \mathbb{N}$$ In particular, I'm pretty much interested in $a=4$ case, but if the general solution ...
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4answers
525 views

Find closed form solution using generation function for the binomial coefficients

I don't have any idea how to start this problem. Could you give a hint? Find closed form solution using generation function for the binomial coefficients: $$a_n:=\sum_{k=0}^{n}\binom{n}{k}^2(-1)^k$$ ...
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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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Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$

I just found this identity but without any proof, could you just give me an hint how I could prove it? $$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$ I know that $$2^n = ...
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426 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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318 views

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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595 views

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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775 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, ...
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209 views

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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How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$

show that $$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$ it's well know that $$(1+\dfrac{1}{n})^n<e$$ so $$(1+\dfrac{1}{16})^{16}<e$$ But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$ ...
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1answer
373 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?
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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 ...
7
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2answers
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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How to prove the following limit: $\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0$?

How to prove the following limit: $$\lim_{n \to +\infty} 4^n\left[\sum_{k=0}^n (-1)^k{n\choose k}\ln (n+k)\right]=0?$$
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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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326 views

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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1answer
186 views

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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216 views

About one generating function

Initially, I have the following problem: find $$\sum_{k=0}^{n+1}(−1)^{n−k}4^k{n+k+1 \choose 2k}.$$ I thought, if I found the function $g_n(x) = \sum_{k=0}^{n}{n+k \choose 2k}x^k$, the answer would be ...
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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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1answer
480 views

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
164 views

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 ...
7
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1answer
209 views

$\sum_{i=0}^m \binom{m-i}{j}\binom{n+i}{k} =\binom{m + n + 1}{j+k+1}$ Combinatorial proof

Is there a simple combinatorial proof for the following identity? $$\sum_{0\leq i \leq m} \binom{m-i}{j}\binom{n+i}{k} =\binom{m + n + 1}{j+k+1}$$ where $m,j \geq 0$, $k \geq n \geq 0$.
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1answer
295 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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2answers
307 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 ...
7
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1answer
162 views

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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3answers
113 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 ...
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1answer
169 views

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 ...