2
votes
2answers
55 views

Alternating sum of binomial coefficients is equal to zero [duplicate]

Prove without using induction that the following formula:$$\sum_{k=0}^n (-1)^k\binom{n}{k}=0$$ is valid for every $n\ge1$. Progress For each odd $n$ we can use the ...
2
votes
2answers
70 views

How to prove $n! > n^a$ for all $a\in \mathbb{R}$ (for sufficiently large $n$)?

I've encountered a proof which claims $n! > n^2$ for sufficiently large $n$. I tried using induction to prove it for an arbitrary $a$: $n! > n^a$. Lets assume the claim is true for $n$: $n! ...
1
vote
2answers
40 views

Problem involving summation and binomial coefficient

I have been fighting with this but I'm really not getting anywhere. $$\sum_{0\leq2k\leq n}\binom{n}{2k}2^k\equiv0\pmod 3$$ $$iff$$ $$n\equiv2\pmod 4$$ Hint: Consider ...
3
votes
3answers
118 views

Evaluation of a sum of $(-1)^{k} {n \choose k} {2n-2k \choose n+1}$

I have some question about the paper of which name is Spanning trees: Let me count the ways. The question concerns about $\sum_{k=0}^{\lfloor\frac{n-1}{2} \rfloor} (-1)^{k} {n \choose k} {2n-2k ...
2
votes
2answers
88 views

Proof of equality $\sum_{k=0}^{m}k^n = \sum_{k=0}^{n}k!{m+1\choose k+1} \left\{^n_k \right\} $ by induction

I have a problem with following equality: $$\sum_{k=0}^{m}k^n = \sum_{k=0}^{n}k!{m+1\choose k+1} \left\{^n_k \right\} $$ And I would like to use induction in following way: Base: $$ m = n $$ And: $$ ...
3
votes
3answers
51 views

Sum of products of binomial coefficients

In a proof I've come across the following identity: $$ \sum_{l=0}^{n-j} \binom{M-1+l}{l} \binom{n-M-l}{n-j-l} = \binom{n}{j} $$ I see that it's right, when plugging in numbers, but I don't see the ...
0
votes
1answer
65 views

Multiple sum involving binomial factors

Let $n$ and $m$ be positive integers and let $0 \le j \le n-m-1$. Show that: \begin{align} \sum\limits_{l=m}^{n-j-1} \binom{n-l-1}{j} \binom{l}{m} \binom{n+l}{j} &=\sum\limits_{p=0}^j ...
2
votes
1answer
73 views

A double sum with combinatorial factors

Let $n$, $p$ and $j$ be integers. As a byproduct of some other calculations I have discovered the following identity: \begin{equation} \sum\limits_{p=0}^{j} \sum\limits_{p_1=0}^j \binom{p+p_1}{p_1} ...
3
votes
1answer
37 views

How to simplify a sum with binomial coefficients multiplied by $k^3/2^k$?

The sum is $$\sum_{k=5}^{\infty}\binom{k-1}{k-5}\frac{k^3}{2^{k}} $$ The first thing I thought of was the binomial coefficient. So I re-indexed it ...
1
vote
3answers
92 views

Simplifying $\displaystyle\sum_{k=0}^{20}(k+4)\binom{23-k}{3}$

In trying to simplify my answer to a problem posted recently, I am trying to show that $\displaystyle\sum_{k=0}^{20}(k+4)\binom{23-k}{3}=8\binom{24}{4}$. I know that ...
1
vote
0answers
53 views

A sum with binomial coefficients in the numerator and denominator.

I am struggling with a combinatorial sum as apart of a long statistical-mechanics derivation. I would appreciate any help. I seek the result of the following summation, for integer $\ell,n$, and ...
0
votes
1answer
25 views

Probability $\sum_{j=n+1}^{2n+1} {M \choose m+1}{M-m-1 \choose j-m-1}/{N \choose j} $

I have a prob. problem: A school has $N$ students in which $M$ students are leader (of each class in school), and $N>M$. There are $2n+1$ balls in the black box including $n+1$ blue balls and $n$ ...
1
vote
1answer
56 views

An upper bound and simplification for expression

I would like to find the upper bound (or simplification) of this expression: $$\sum_{j=1}^{n+1}\sum_{i=0}^{j-1} a^{j+i} {j+i \choose i}{n+1\choose j}{n \choose i}/{2n+1 \choose j+i}$$ where ...
1
vote
2answers
158 views

How find this sum $\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$

Find the sum close form $$f(x)=\sum_{i=0}^{2n}\dfrac{\binom{2n}{2i}\binom{2i}{i}x^{2i}}{2^{2i}}$$ if we let $$\dfrac{x}{2}=y$$ then $$f(y)=\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$$ ...
0
votes
2answers
63 views

Simplify the expression of binom

Any one knows how to simplify this expression or finding upper bound of this expression: $$\sum_{j=1}^{n+1}\sum_{i=0}^{j-1} a^{j+i} {j+i \choose i}$$ where $0<a<1$ is constant. Thanks a lot.
1
vote
1answer
30 views

Upper bound of $\sum\limits_{j=1}^{n+1} \sum\limits_{i=0}^{j-1}{n+1 \choose j}{n \choose i}$

I would like to find max (or sup.) of the sum: $$S=\sum\limits_{j=1}^{n+1} \sum\limits_{i=0}^{j-1}{n+1 \choose j}{n \choose i}.$$ I found $S\le \frac{1}{\sqrt{\pi n}}.2(n+1).4^n$ but It seems it's ...
5
votes
3answers
85 views

Ordered partitions of an integer (with a twist)

I would like to know how to prove (preferably algebraically) that $P_1(2,n)=F_{2n+1}$, where $P_1(2,n)$ is what I define to be the number of ordered partitions of an integer, where the number $1$ has ...
3
votes
1answer
72 views

Upper bound of $S=\sum_{k=1}^{P}k!\binom{P}{k}\binom{Q}{k}$

EDIT: How can I find a good upper bound to this quantity ? $$S_{n,m}=\sum_{k=1}^{P}k!\binom{P}{k}\binom{Q}{k}$$ where $P=\min\{m,n\}$ et $Q=\max\{m,n\}$.
7
votes
6answers
166 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 ...
3
votes
2answers
87 views

Prove by induction that $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$ is decreasing

I want to prove that the following sequence is monotonously decreasing: $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$ I think it should be ...
2
votes
0answers
49 views

Sum and binomials

I have this sum ...
6
votes
2answers
170 views

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. ...
1
vote
2answers
79 views

Partial sum of binomial

I 'm trying to figure out a closed form solution for the following summation: $\sum_{j=0}^{\omega} j{n \choose j}p^{j}(1-p)^{n-j}$ where $\omega < n$ Is there any closed form solution?
9
votes
3answers
161 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..?
1
vote
0answers
39 views

Sum involving binomial coefficients

Exist a closed form for $$\left(-1\right)^{N}\underset{i=1}{\overset{N}{\sum}}\left(-1\right)^{i}\dbinom{N}{i}\dbinom{N+i}{i-1}\,\frac{1}{2i+1}?$$ I think I've to use in some way the formula of the ...
4
votes
0answers
142 views

Proving an equation involving binomial coefficients

Prove that $$\sum_{q=0}^v \binom{v}{q}\frac{q!}{v^{q+1}} = \sum_{q=0}^{v-1} \binom{v-1}{q} \frac{(q+2)!}{v^{q+2}}$$ Thanks. Below are what I have tried: Approach 1: $$\sum_{q=0}^{v-1} ...
2
votes
1answer
47 views

Identity with binomials [duplicate]

Does there exist a closed formula for $$\underset{n=1}{\overset{N-1}{\sum}}\dbinom{N+n}{n}?$$ I've searching on wikipedia but I haven't found this kind of sum.
2
votes
0answers
63 views

Sum of product of binomial coefficients and exponential function

I would like to know how to obtain (if it exists) a closed form expression of the sum $$S=\sum^{n}_{k=0}2^k{{n+1}\choose k}{{r-n-2}\choose {n-k}}$$ So far, I have tried to use the method of ...
0
votes
0answers
24 views

Simplification of a power weighted alternating binomial sum

Given positive integers $T$, $n$ and $m$ and real number $p$ with $0< p < 1$, how can I simplify the following binomial sum: $$ \sum_{k=m}^{\lfloor ...
1
vote
1answer
20 views

Proving an identity involving binomial coefficients and fractions

I've been trying to prove the following formula (for $n > 1$ natural, $a, b$ non-zero reals), but I don't know where to start. $$\sum_{j=1}^n \binom{n-1}{j-1} \left( \frac{a-j+1}{b-n+1} \right) ...
3
votes
2answers
73 views

Integer sum as binomial coefficient

What's the rule for expressing integer sums as binomial coefficients? That is, for $p=1$ it is $$\sum_{n=1}^N n^p = {{N+1}\choose 2} $$ What is it for higher powers?
1
vote
1answer
24 views

A sum of Laguerre polynomials

I'm looking to find a closed-form expression for the sum $$S = \sum_{n=0}^N e^{-x/2} L_n^{0}(x),$$ where $L_n^{0}$ is the $n$th Laguerre polynomial. Using the formula $$L_n^{\alpha}(x) = \sum_{m=0}^n ...
3
votes
4answers
117 views

Proving Combinatorical Summation: $n!=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n$ [duplicate]

been stuck with this question for the last few hours, any help would be appreciated. $$ {\large n! = \sum_{k = 0}^{n}\left(-1\right)^{k}{\,n\, \choose \,k\,} \left(\,n - k\,\right)^{n}} $$ what I ...
0
votes
0answers
64 views

What's the interpretation of $\sum_{i,j} i \cdot j \cdot \binom{2n}{i}\cdot \binom{2n}{j} \cdot \binom{2n}{3n-i-j}$?

I'm having problems with finding the combinatorial interpretation of this sum: $$\sum_{i,j} i \cdot j \cdot \binom{2n}{i}\cdot \binom{2n}{j} \cdot \binom{2n}{3n-i-j}$$ Can anyone help, please?
1
vote
1answer
29 views

How find this sum $\sum_{j=0}^{\infty}\binom{m+2j}{m}t^{2j},0<t<1$

Let $m$ is give postive integer numbers, Find the sum $$\sum_{j=0}^{\infty}\binom{m+2j}{m}t^{2j},0<t<1$$ if this not have closed form,and can you use Special function ?
0
votes
1answer
39 views

Is it possible to get a formula for this summation

The binomial sum $$s_n=\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots+\binom{2n}{n}$$ I tried solving through recurrence, but unable to find one.
2
votes
1answer
67 views

simplifying a triple sum of products of binomial coefficients

Right now I have a horribly-looking triple sum ($x,y,z$ are non-negative integers and $x+y+z=N$): $$ W_{12}(x,y)=\frac{x}{N}\sum_{l=0}^{x-1}\sum_{l'=0}^{y}\sum_{l''=0}^{z}{x-1 \choose ...
2
votes
1answer
114 views

Combinatorial proof involving reciprocals

This is a follow-up to this question: show that if $n$ is a positive integer then $$\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}\binom{n}{k} =\sum_{k=1}^{n}\frac{1}{k}\ .$$ I was able to answer the question by ...
8
votes
4answers
210 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. $S_n$ ...
5
votes
1answer
90 views

An inverse binomial summation.

I am looking for a closed form for this summation: $$ \sum_{j=1}^m\frac{r^{-j}}{j{m\choose j}} = \sum_{j=1}^m\frac{r^{-j}}{m{m-1\choose j-1}} = \frac1{rm} \sum_{k=0}^{m-1}\frac{r^{-k}}{{m-1\choose k}} ...
3
votes
0answers
117 views

Proving $\sum_{k=1}^{n}\binom{n-1}{k-1}{\binom{n+k}{k}}^{-1}=\frac 12$ combinatorially

Question : How can we prove the following equations combinatorially? $$\begin{eqnarray}\sum_{k=1}^{n}\frac{\binom{n-1}{k-1}}{\binom{n+k}{k}}&=&\frac ...
0
votes
2answers
70 views

Evaluate the sum $\sum_{0\leq j < k\leq n}\binom{n}{j}\binom{n}{k}$

Could someone give me a hint on how to do this? I believe I know what the answer to be (I computed some low values and checked on OEIS). However, I was hoping someone would be able to explain to me ...
4
votes
2answers
81 views

Evaluating Combination Sums

Evaluate $$\sum_{k=0}^n{n+k\choose 2k} 2^{n-k}$$ So im not really sure how to begin with this. I would imagine we start with dividing out $2^{n}$, but not really sure much past that
11
votes
2answers
293 views

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 with playing numbers, I conjectured ...
1
vote
2answers
36 views

Proving this binomial identity

I'm required to prove the following binomial identity: $$\sum\limits_{k=0}^l {n \choose k} {m \choose l-k} = {n+m \choose l}$$ I tried various arrangements but reached nowhere. Finally I turned to ...
1
vote
0answers
35 views

How find the sum $2\sum_{k=1}^{\infty}\sum_{i=0}^{2k-1}\frac{\binom{2k}{i}\cdot B_{i}\cdot(m-1)^{2k-i}}{2k(2k-1)m^{2k-1}}$

Find the sum $$2\sum_{k=1}^{\infty}\sum_{i=0}^{2k-1}\dfrac{\binom{2k}{i}\cdot B_{i}\cdot(m-1)^{2k-i}}{2k(2k-1)m^{2k-1}}$$ where $B_{i}$ is Bernoulli numbers. my idea: since ...
0
votes
1answer
58 views

Show $\large\sum\limits_{j=0}^{r}\binom{j+k-1}{k-1}=\binom{r+k}{k}$

Show $\large\sum\limits_{j=0}^{r}\binom{j+k-1}{k-1}=\binom{r+k}{k}$ Hint: Place $r$ balls in $m$ urns, in how many of this arrangements can you find $b$ balls in the first urn. I'm sure that ...
6
votes
5answers
182 views

Formula for $\sum_{k=0}^n k^d {n \choose 2k}$

If $d \geq 1$ is an integer, is there a general formula for $$\sum_{k=0}^n k^d {n \choose 2k}\,?$$ We know that $\sum_{k=0}^n k {n \choose 2k} = \frac{n2^n}{8}$ and $\sum_{k=0}^n k^2 {n \choose 2k} = ...
4
votes
3answers
211 views

How to closed the sum $\displaystyle \sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$

How to closed the sum $\displaystyle S=\sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$ I'm trying divide two cases $n$ odd and $n$ even. I predict that ...