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

Please provide a combinatorial proof for the following: Prove the identity $$\sum_{i=0}^{k}{m+k-i-1 \choose k-i}{n+i-1 \choose i}={m+n+k-1 \choose k}$$ Hint: use idea of "selection with repetition". ...
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Proof that binomial coefficient is a natural number [duplicate]

Possible Duplicate: Proof that a Combination is an integer What is the proof that the binomial coefficient is a natural number? $$k\ge0,n\ge k \implies {n \choose k} \in N,$$ I guess ...
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Maximum term of (a + b) ^ n

I would like a demonstration of the fact below. Being given real numbers a and b (nonzero) and a positive integer n, the order p, that occupies the maximum term (in absolute value) of the ...
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Another sum involving binomial coefficients.

Let both $a$,$b$ and $\theta$ be real numbers not equal to a negative integer. Let $m$ be a positive integer. I have shown that the following equality holds: \begin{eqnarray} ...
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Prove Maximum term in the expansion.

How should I prove the maximum term in the expansion of $(x+a)^n$ where $ax>0$ is the term $C(n,r)x^{(n-r)}a^r$ for which $r= \left[\cfrac{(n+1)}{(n/a)+1} \right]$ ?
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Proof of a binomial identity $\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$

Prove that $$\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$$ The exercise provides the following hint: $\,\,\displaystyle{n \choose k}={n\choose n-k}$. Any help?
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Solutions to $\binom{n}{5} = 2 \binom{m}{5}$

In Finite Mathematics by Lial et al. (10th ed.), problem 8.3.34 says: On National Public Radio, the Weekend Edition program posed the following probability problem: Given a certain number of ...
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X'mas Combinatorics

Inspired the various** algebraic X'mas greetings sent to me over the festive period, I thought I would try to devise one of my own. $$\Large ...
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Proofs of $\lim\limits_{n \to \infty} \left(H_n - 2^{-n} \sum\limits_{k=1}^n \binom{n}{k} H_k\right) = \log 2$

Let $H_n$ denote the $n$th harmonic number; i.e., $H_n = \sum\limits_{i=1}^n \frac{1}{i}$. I've got a couple of proofs of the following limiting expression, which I don't think is that well-known: ...
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Exercise from Comtet's Advanced Combinatorics: prove $27\sum_{n=1}^{\infty }1/\binom{2n}{n}=9+2\pi \sqrt{3}$

In exercise 36 Miscellaneous Taylor Coefficients using Bernoulli numbers on pages 88-89 of Louis Comtet's Advanced Combinatorics, 1974, one is asked to obtain the following explicit formula for the ...
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Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$

Recently, I was able to prove that $$\sum_{n=1}^\infty \frac{1}{n \binom{2n}{n}}= \frac{\pi}{3\sqrt{3}}$$ $$\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}}= \frac{\pi^2}{18}$$ But does anybody know ...
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The Hexagonal Property of Pascal's Triangle

Any hexagon in Pascal's triangle, whose vertices are 6 binomial coefficients surrounding any entry, has the property that: the product of non-adjacent vertices is constant. the greatest common ...
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How to prove $ \sum_{k=0}^n \frac{(-1)^{n+k}{n+k\choose n-k}}{2k+1}=\frac{-2\cos\left(\frac{2(n-1)\pi}{3}\right)}{2n+1}$

How to prove $$\sum_{k=0}^n \binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=-\frac{2}{2n+1}\,\cos\left(\frac{2(n-1)\pi}{3}\right)\;\text{?}$$ I have a proof by induction for it, but it isn't simple! I want ...
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A Combinatorial Proof of Dixon's Identity

Dixon's Identity states: $$ \sum_{k} (-1)^k\binom {a+b}{b+k}\binom{b+c}{c+k}\binom{c+a}{a+k} = \binom{a+b+c} {a,b,c}$$ A bit of history: The case $a=b=c$ was proved by Dixon in 1891 using ...
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Limit of $\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1}\right)$

I'm trying to calculate the limit for the sum of binomial coefficients: $$S_{n}=\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1} \right).$$
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What is the binomial sum $\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n}$ in terms of zeta functions?

We have the following evaluations: $$\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = ...
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Strehl identity for the sum of cubes of binomial coefficients

In 1993 Strehl showed that $$ \sum_k\binom nk^3=\sum_k\binom nk^2\binom{2k}n. $$ I’m interested in a combinatorial proof. Upd (Jan '14). Maybe the original question was too restrictive — I'm now ...
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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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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? ...
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proof of formula and calculation sum

Show that following formula is true: $$ \sum_{i=0}^{[n/2]}(-1)^i (n-2i)^n{n \choose i}=2^{n-1}n! $$ Using formula calculate $$ \sum_{i=0}^n(2i-n)^p{p \choose i} $$
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Is there a direct proof of this lcm identity?

The identity $\displaystyle (n+1) \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \text{lcm}(1, 2, ... n+1)$ is probably not well-known. The only way I know how to prove ...
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Binomial Sum Related to Fibonacci: $\sum\binom{n-i}j\binom{n-j}i=F_{2n+1}$

How would I prove $$ \sum\limits_{\vphantom{\large A}i\,,\,j\ \geq\ 0}{n-i \choose j} {n-j \choose i} =F_{2n+1} $$ where $n$ is a nonnegative integer and $\{F_n\}_{n\ge 0}$ is a sequence of ...
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4answers
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Expected Value of a Binomial distribution?

If $\mathrm P(X=k)=\binom nkp^k(1-p)^{n-k}$ for a binomial distribution, then from the definition of the expected value $$\mathrm E(X) = \sum^n_{k=0}k\mathrm P(X=k)=\sum^n_{k=0}k\binom ...
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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 ...
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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 ...
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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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Prove $(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j} $

I stumbled upon the identity $$(1-x)^{2k+1} \sum\limits_{n\ge 0}\binom{n+k-1}{k}\binom{n+k}{k} x^n = {\sum\limits_{j\ge 0} \binom{k-1}{j-1}\binom{k+1}{j} x^j}. $$ The right-hand side is a polynomial. ...
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Binomial Congruence

How can we show that $\dbinom{pm}{pn}\equiv\dbinom{m}{n}\pmod {p^3}$ for positive integers m and n and p a prime greater than 5? I can do it for mod p^2 but Im stuck here.
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Are these two binomial sums known? Proven generalization to the Hockey Stick patterns in Pascal's Triangle

English translation. You can see the original - deprecated - in Portuguese here Hi, I arrived at a generalization for the Hockey Stick Patterns, from our beloved Pascal's Triangle. This ...
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Closed form for $\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$

How can I calculate the following sum involving binomial terms: $$\sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{(k+1)^2}$$ Where the value of n can get very big (thus calculating the binomial ...
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How prove this $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1 $

Show that $$\sum_{k=0}^{n}\dfrac{\binom{2n-k}{n}}{2^{2n-k}}=1$$ I think this problem can be solved with nice methods, such as algebraic ones. Or can I use probability methods? Thank you
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A conjecture including binomial coefficients

Question: $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}$$ is true for every $n\in \mathbb N$? If this is true, then how can we prove this? When I was playing with numbers, I conjectured ...
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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: $$\sum_{k=0}^{\min[p,q]}{p\choose k}{q\choose k}{n+k\choose p+q}={n\choose p}{n\choose q}.$$
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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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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, ...
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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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Truncated alternating binomial sum

It is easily checked that $\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0$, for example by appealing to the binomial theorem. I'm trying to figure out what happens with the ...
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Summation of weighted squares of binomial coefficients

Show that $$\sum_{k=0}^n \left[ \frac{n-2k}{n} {n\choose k}\right]^2=\frac{2}{n}{2n-2 \choose n-1}.$$
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Spivak's Calculus - Exercise 4.a of 2nd chapter

4 . (a) Prove that $$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}.$$ Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$. I'm having a hard time trying to solve the problem ...
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Combinatorial proof for $\binom{n}{a}\binom{a}{k}\binom{n-a}{b-k} = \binom{n}{b}\binom{b}{k}\binom{n-b}{a-k}$?

I have to prove the following using a combinatorial proof: $\binom{n}{a}\binom{a}{k}\binom{n-a}{b-k} = \binom{n}{b}\binom{b}{k}\binom{n-b}{a-k}$ Ok, so here is what I have worked out so far: We ...
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Hard binomial sum [closed]

How to prove this relation? $$\sum_{i=0}^{n}\frac{2^{-2i}\binom{2i}{i}}{n+i+2}=\frac{2^{4n+2}-\binom{2n+1}{n}^2}{(2n+3)2^{2n+1}\binom{2n+1}{n}}$$ That seems difficult!
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Evaluate a sum with binomial coefficients

$$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$ I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$ ...
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Fitting a closed curve on the roots of ${x \choose k}-c$

Let $${n \choose k} = \frac1{(n+1) \operatorname{B}(n-k+1, k+1)}.$$ be the generalized binomial coefficient, and here $\operatorname{B}$ is the Beta function. Let $f_{k,c}(x)$ be the following ...
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Asymptotics of the sum of squares of binomial coefficients

We are trying to estimate the cardinality $K(n,p)$ of so-called Kuratowski monoid with $p$ positive and $n$ negative linearly ordered idempotent generators. In particular, we are interesting in the ...
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Proving identities like $\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}$ combinatorially

I have to give a combinatorial proof of $$\sum_{k=1}^nk{n\choose k}^2=n{2n-1\choose n}.$$ I find it difficult to solve such problems. I'm not a brilliant person and never will be so I need to have ...
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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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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. ...
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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 ...
7
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Roots of a polynomial whose coefficients are ratios of binomial coefficients

Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation $$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + ...