All Questions
984
questions
0
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1
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Binomial identity involving square binomial coefficient [closed]
I want to prove this identity, but I have no idea...
Could someone please post a solution? Thank you.
$$\sum_{k=0}^{n} \binom{-1/2}{n+k}\binom{n+k}{k}\binom{n}{k}= \binom{-1/2}{n}^2$$
(Maybe -1/2 can ...
2
votes
2
answers
91
views
Compute the value of a double sum
I need some help computing a(n apparently nasty) double sum:
$$f(l):=\sum_{j = \frac{l}{2}+1}^{l+1}\sum_{i = \frac{l}{2}+1}^{l+1} \binom{l+1}{j}\binom{l+1}{i} (j-i)^2$$
where $l$ is even. I'm not ...
3
votes
3
answers
65
views
Prove that $\frac{(n + 1)!}{((n + 1) - r)!} = r \sum_{i=r - 1}^{n} \frac{i!}{(i - (r - 1))!}$
I Need Help proving That
$$\frac{(n + 1)!}{(n - r + 1)!} = r \cdot \sum_{i=r - 1}^{n} \frac{i!}{(i - r + 1)!}$$
Or in terms of Combinatorics functions:
$P_{r}^{n+1} = r \cdot \sum_{i = r-1}^{n} {P_{r-...
3
votes
0
answers
162
views
Closed form for $\sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$?
I've found this sum:
$$S(n) = \sum_{k=n}^{2n-1} k(-1)^k \sum_{i=k}^{2n-1} {2n \choose i+1} {i \choose n}$$
The inner sum is elliptical iirc, but perhaps the double sum has a nice expression. We can ...
2
votes
2
answers
47
views
Given a set of integers, and the number of summations find the resulting frequencies
Given a set $X = \{x_1,x_2,...x_m\}\subset \mathbb{Z}$ and the number of allowed addends $N$. How can I find the frequency of every possible sum?
Example: $X = \{-1, 2\}$ and $N=3$ then every ...
0
votes
0
answers
45
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Closed form for nested sum involving ratios of binomial coefficients
I ran into the following nested sum of binomial coefficients in my research, but I couldn't find the closed form expression for it. I looked at various sources and still couldn't find the answer. So I ...
3
votes
1
answer
84
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Combinatorial Proof of the $\sum_{k=1}^{n} k^2 \binom{n}{k} = n(n + 1) 2^{n - 2}$ [duplicate]
How to prove that $$\sum_{k = 1}^{n}k^2\binom{n}{k} = n(n + 1)2^{n -2}$$ in an combinatorial way?
I have a algebraic proof method, but I don't know how to use a combinatorial proof method to do it.
0
votes
1
answer
128
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Spivak Exercise, Prove Vandermonde's Identity $\sum_{k=0}^{l}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}$ [duplicate]
Prove that
$$\sum_{k=0}^{l}\binom{n}{k}\binom{m}{l-k}=\binom{n+m}{l}$$
Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$
Proof:
Following Spivak's advice, we have
$\sum_{k=0}^{n}\binom{n}{k}x^k=(...
0
votes
0
answers
44
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Choosing an ordered triplet of non-negative integers $(m_1, m_2, m_3)$ such that $m_1 + m_2 + m_3 = n$.
Define
$$A = \{ (m_1, m_2, m_3) : m_1 \geq 0, m_2 \geq 0, m_3 \geq 0, m_1 + m_2 + m_3 = n\}$$
Given that $n \geq 0$ and $n, m_1, m_2, m_3 \in \mathbb{Z}^+$, then why it is the case that
$$\vert A \...
0
votes
2
answers
45
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Trying to prove equivalence of combinatorial formula and nested summations
I’m sorry if this is a dumb problem, but I’m trying to get into mathematics and prove this but I’m only in 9th grade and haven’t found any sources on this:
Prove that $$ {n \choose r} = \overbrace{ \...
4
votes
1
answer
78
views
Identity regarding the sum of products of binomial coefficients.
Consider the following toy problem
Person A and Person B have $n$ and $n+1$ fair coins respectively. If they both flip all their coins at the same time, what is the probability person B has more ...
2
votes
1
answer
69
views
Closed form for a sum of binomial coefficients
Let $m,n,r\in\mathbb{N}\cup\{0\}.$ I am interested in finding a closed form for the sum
$$\sum_{i=0}^m{{n+i}\choose{r+i}}.$$
Let $f(m,n,r)$ denote the above sum.
We may make a few trivial observations....
1
vote
2
answers
92
views
Alternating sum involving binomial coefficients
I want to prove that
$$
\sum_{i=0}^{n}{n\choose i}
\frac{\left(1 + \alpha i\right)^{n}
\left(-1\right)^{n - i}}{n!} = \alpha^{n}.
$$
This is a guess based on the computations for
$n = 0,1,2,3$.
Do you ...
1
vote
3
answers
146
views
Evaluate: $\sum_{i=1}^{\lfloor (n+1)/2\rfloor}i\binom{n-i+1}{i}$
Is there a closed form of the expression
$$
\sum_{i = 1}^{\left\lfloor\left(n + 1\right)/2\right\rfloor}
i\binom{n - i + 1}{i}
$$
My Attempt:
From what I observe it is a situation where
there are $n/...
4
votes
0
answers
128
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Calculating $\sum\limits_{k=0}^n\binom{n}{k}/\left(2^k+2^{n-k}\right)$
I am trying to find a closed form for $\sum\limits_{k=0}^n\frac{\binom{n}{k}}{2^k+2^{n-k}}$. I saw on quora that integration can be used to rewrite portions of such equations, and so I attempted this. ...