Questions tagged [binomial-coefficients]

For questions involving the coefficients involved in the binomial theorem. $ \binom{n}{k}$ counts the subsets of size $k$ of a set of size $n$.

The binomial coefficient $\binom{n}{k}$ can be defined in several equivalent ways for $n$ and $k$ non-negative integers:

  1. The number of subsets of size $k$ of a set of size $n$.
  2. Element $k$ of row $n$ in Pascal's triangle (counting the first element or row as $0$).
  3. $\dfrac{n!}{k!(n-k)!}$
  4. The coefficient of $x^k$ in $(1+x)^n$.

The binomial theorem says that $$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k$$ using the convention that $0^0=1$.

Binomial coefficients can be extended for arbitrary complex $\alpha$ through the formula: $$\binom{\alpha}{k}=\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-k+1)}{k(k-1)(k-2)\dots1}$$

7695 questions
1
vote
1 answer

Is it possible to evaluate this binomial sum?

Would it be possible to evaluate this sum? $$\sum_{k=0}^{N/2}k\binom{N+1}{k},$$ where $N$ is even? I know that the sum $$\sum_{k=0}^{N+1}k\binom{N+1}{k}=2^N(N+1)$$ (by http://mathworld.wolfram.com/BinomialSums.html, equation $(21)$), but I can't…
1
vote
1 answer

An identity involving the binomial theorem

While working on another problem that involved a linear system of arbitrary size $n$, I managed to empirically come up with a solution for that system. My solution is correct if and only if the following identity is true: $$\sum_{i=1}^n(-1)^i…
Kyle
  • 113
1
vote
4 answers

Determine the coefficient

I've been stuck on this question for a little while, could someone point me in the right direction? I'm supposed to determine the coefficient of $x^8$. $$x^8\quad in \quad \frac{x}{(1-x)(1-2x)} $$ I know I have to use binomial theorem however, I'm…
1
vote
3 answers

Calculate sum $S=\sum_{k=0}^{n}\binom{k}m$

Calculate sum $$S=\sum_{k=0}^{n}\begin{pmatrix} k \\ m\end{pmatrix}$$ My solution if $n
Bosen
  • 21
1
vote
1 answer

The biggest binomial coefficient in $(n+\frac 1n)^n$ if the product of the fourth member in the expansion and the fourth from back member is 14400

I'm stuck on this one. I am not expected to know how to solve cubic equations so this gets even more confusing, as i get $\binom n 3 = 120$. So I can't even calculate $n$. Is there a way to go around solving the cubic equation?
John Doe
  • 1,080
1
vote
1 answer

What's the solution to this binomial?

what's the coefficient of $x^6$ in the expansion of $(1+X^2+X)^{-3}$? I have factorized the term to $\left(\frac{1-x^{-3}}{1-x}\right)^3$ after this I'm having problem solving it
Aman Bisht
1
vote
2 answers

How can I prove this property of binomial coefficients?

I was playing around with binomial coefficients and binomial expansions and I came across an interesting identity: $$ \sum_{k=0}^n\frac{1}{k+1}{n \choose k}x^k=\frac{(x+1)^{n+1}-1}{(n+1)x}$$ I have verified this for a few different values of n and…
ASKASK
  • 9,000
  • 4
  • 28
  • 49
1
vote
3 answers

How to show $u_n=\binom{n}{l}\Rightarrow s_n=\binom{n+1}{l+1}$?

Hello respected everyone. Before I ask my query, let us first define binomial coeffcients as follows: For $n, r\in \mathbb N$, we define $$\binom{n}{r}=\begin{cases} \frac{n!}{r!(n-r)!}~~~ \text{if} ~~~0\leq r\leq n\\ 0~~~~~~~~~~~~~…
KON3
  • 4,111
1
vote
1 answer

Show that $\binom{1/2}{k} = \frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}$

The Problem Show that $$\binom{1/2}{k} = \frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k}$$ My Work $$\begin{align*}\frac{(-1)^{k+1}}{4^k(2k-1)}\binom{2k}{k} &= \frac{(-1)^{k+1}(2k)!}{4^k(2k-1)(k!)^2}\\\\ &=…
Dunka
  • 2,787
  • 12
  • 41
  • 69
1
vote
1 answer

Binomial series after using binomial identity

Following http://en.wikipedia.org/wiki/Binomial_coefficient#Newton.27s_binomial_series , I am trying to prove that $$ \sum_{\kappa=0}^\infty \binom{\eta + \kappa}{\kappa}x^\kappa = (1 - x)^{-(\kappa+1)}. $$ According to the article, this identity…
nomen
  • 2,707
1
vote
1 answer

Express $\binom{n+2}{k}$ according to $\binom{n}{k}$

I've just begun studying binomial coefficient and I'm trying to express $\dbinom{n+2}{k}$ according to $ \dbinom{n}{k}$. With this result I have to conclude that $\dbinom{2n}{2k}, \dbinom{2n+1}{2k}$ and $\dbinom{2n+1}{2k+1}$ are even if and only if…
Tom
  • 27
1
vote
2 answers

Binomial Expansion where N is negative

Comparing the formula for regular binomial expansion (n>1): $(a+b)^n=a^n + \binom{n}1a^{n-1}b + \binom{n}2a^{n-2}b^2 +...$ to binomial expansion for negative indices, (n<1): $(1+x)^n= 1 + nx + \dfrac{n(n-1)x^2}{2!} +...$ Does this imply that…
user61871
  • 197
1
vote
1 answer

What is the smallest value beside 1 of a binomial with two integer values > 0?

I'm searching for the smallest possible value of a binomial(a, b) where a >= b and both values are greater than 0. I calculated a few binomials and always got the result a as the smallest possible value. Is this always the case or just a…
xxsl
  • 99
1
vote
5 answers

$C(n,p) = C(n,q) => p + q= n$

How to prove the following theorem? Theorem: If the binomial numbers $C(n,p)$ and $C(n,q)$ are equal, with $p$ different from $q$, then $p+q = n$. ($n$, $p$ and $q$ are natural numbers)
Paulo Argolo
  • 4,210
1
vote
3 answers

${n\choose m}={n\choose n-m}$ Proof

I need to prove the following: ${n\choose m}={n\choose n-m}$ With the definition: ${n\choose m}= \left\{ \begin{array}{ll} \frac{n!}{m!(n-m)!} & \textrm{für \(m\leq n\)} \\ 0 & \textrm{für…
Petra
  • 73