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2 votes
0 answers
77 views

Closed expression for a combinatorial sum

The following equality is true for every positive integer $n$ : $$\sum_{k=0}^n {n \choose k} = 2^n $$ It is a special case ($p = 0$) of the sequence : $$S_{p, n}=\sum_{k=0}^n k^p {n \choose k} $$ For ...
alati ahmad's user avatar
2 votes
1 answer
342 views

Polynomial representation of $\sum\limits_{n=m+1}^N \binom {n-1} m$

Background: I am looking for a polynomial representation of $$ \sum_{n=m+1}^N \binom {n-1} m \tag{1} $$ where $m\in\mathbb{Z^+}$ is a positive integer $2,3,4,\ldots $ that is greater than or equal to $...
Axion004's user avatar
  • 10.1k
1 vote
1 answer
114 views

A fun double sum involving binomial coefficients

I came across the following double sum expression: $$ S(k) := \sum_{l=0}^{2(k+1)-1} \sum_{i=0}^{l} (-1)^i\binom{2(k+1)-i}{2(k+1)-l} \left[ 3\binom{2k}{i-2}+ 3k\left(\binom{2k}{i-1} - \binom{2k-1}{i-2}\...
ipreferpi42's user avatar
3 votes
1 answer
75 views

How to prove that this binomial sum remains positive for $x>1$?

Let's say you have this function for real numbers $x>1$, for some positive integer $n \geq 1$ $$ \sum_{k=0}^{\left \lfloor n/2 \right \rfloor} {x \choose 2k+\frac{1-(-1)^n}{2}} $$ How would you ...
Baklava Gain's user avatar
1 vote
1 answer
89 views

Asymtotic of some binomial sum

Assume $n$ is a positive odd integer, I need to find the asymptotic as $n$ goes to infinity of the sum $$s(n,x)=\frac1x\sum_{k=0}^n (-1)^k\binom{-x-\frac12}{k}\binom{x-\frac12}{n-k},$$ where the ...
lomarkly's user avatar
  • 134
5 votes
2 answers
169 views

Prove that $\sum_{k=1}^n\frac{\prod_{1\leq r\leq n, r\neq m}(x+k-r)}{\prod_{1\leq r\leq n, r\neq k}(k-r)}=1$

For arbitrary $x$ and $1\leqslant m\leqslant n$, prove the following: $$\sum_{k=1}^n\frac{\prod_{1\leq r\leq n, r\neq m}(x+k-r)}{\prod_{1\leq r\leq n, r\neq k}(k-r)}=1$$ I'm looking for a proof that ...
Rusurano's user avatar
  • 848
1 vote
1 answer
97 views

Sum with binomial coefficient using identity

I want to prove: $\displaystyle \sum_{k=0}^n (-1)^k \binom{x}{k} = (-1)^n \binom{x-1}{n}$ using: $(1-z)^x \cdot \frac{1}{1-z} = (1-z)^{x-1}$ I know how to do it with induction but i somehow can't ...
elson1608's user avatar
  • 121
1 vote
1 answer
33 views

Simplify $\sum_{t=k}^{n} (\binom{n}{t} \cdot a^{t-1} \cdot (1 - a)^{n - t - 1} \cdot (t - n \cdot a))$

I was working on my probability theory homework and I found probability density function that looks as following $$\sum_{t=k}^{n} \binom{n}{t} \cdot a^{t-1} \cdot (1 - a)^{n - t - 1} \cdot (t - n \...
math-traveler's user avatar
6 votes
2 answers
121 views

Is the sum $\sum_{k=0}^n{n \choose k}\frac{(-1)^{n-k}}{n+k+1}$ always the reciprocal of an integer $\big(\frac{(2n+1)!}{(n!)^2}\big)$?

Denote the sum $$S_n := \sum_{k=0}^n{n \choose k}\frac{(-1)^{n-k}}{n+k+1}$$ This value arose in some calculations of polynomial coefficients. I'm not used to dealing with expressions of this sort. ...
Somatic Custard's user avatar
6 votes
4 answers
258 views

If $P(x)$ is any polynomial of degree less than $n$, show that $\sum_{j=0}^n (-1)^j\binom{n}{j}P(j)=0$. [duplicate]

If $P(x)$ is any polynomial of degree less than $n$, then prove that $$\sum_{j=0}^n (-1)^j\binom{n}{j}P(j)=0$$ My approach was to try and prove this separately for $j^k\ \ \forall\ \ k<n$, instead ...
Pravimish's user avatar
  • 641
0 votes
2 answers
51 views

The Polynomials $P_n(x+y)$

Let $$\displaystyle P_n(x) = \sum_{k=0}^n \binom{n}{k}x^k.$$ We need to show that $$P_n(x+y) = \sum_{k=0}^n\binom{n}{k}P_k(x)y^{n-k}.$$ In the proof, we have $$\begin{array}{rcl} P_n(x+y) &=&...
Jerome's user avatar
  • 11
32 votes
4 answers
2k views

Questions on a self-made theorem about polynomials

I recently came up with this theorem: For any complex polynomial $P$ degree $n$: $$ \sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) = 0\quad \forall a,b \in\mathbb{C}$$ Basically, if $P$ is ...
user avatar
3 votes
0 answers
108 views

Is there a simple formula for this polynomial sum?

The following seemingly arbitrary polynomial has popped up in my research on period polynomials of modular forms, and I would like to try to understand it a little better. I hope it is OK to ask about ...
Alex Saad's user avatar
  • 3,559
2 votes
2 answers
119 views

An Identity for a Fibbonacci-Type Polynomial

Problem: The polynomials $p_{n}\left(x\right)$ are defined recursively by the linear homogenous order 2 difference equation $$p_{n+1}\left(x\right)=2\left(1-2x\right)p_{n}\left(x\right)-p_{n-1}\left(...
PalmTopTigerMO's user avatar
2 votes
1 answer
116 views

Coefficients of a polynomial

Consider the polynomial $$ P_n(t)=(-1)^n (2t-1) \frac{(2t-1)^{n-1}-t^{2n-2}}{(t-1)^2},$$ that is up to a factor $(-t^2)^{n-1}$ the Tutte polynomial $\frac{x^n-x}{x-1}$ of the cyclic graph evaluated in ...
Roberto Pagaria's user avatar

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