Question

Prove that $\binom{2n}{n} > \prod_{p > n}^{p<2n}{p}$

score 294 · Answer 1 · 2016-07-30 17:42:06

294

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Prove that $\binom{2n}{n} > \prod_{p > n}^{p<2n}{p}$

Published on 30 Jul 2016 - 17:42

score 202 · Answer 2 · 2026-03-29 10:36:15

202

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Help on how to prove that $\ln{k\choose k/2}=\sum_{n=1}^{\infty}(-1)^{n-1}\left({k\over n}-\ln{n+k\over n}\right)$

Published on 29 Mar 2026 - 10:36

score 941 · Answer 3 · 2016-07-31 18:00:52

941

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Finding the coefficient of $x^r$ in an expansion.

Published on 31 Jul 2016 - 18:00

score 157 · Answer 4 · 2026-04-11 08:45:55

157

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Binomial identity: Clever idea for a short proof?

Published on 11 Apr 2026 - 8:45

score 522 · Answer 5 · 2026-03-25 11:18:07

522

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Generalization of the q-Binomial Theorem

Published on 25 Mar 2026 - 11:18

score 1.6k · Answer 6 · 2016-08-01 15:37:26

1.6k

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What does "n over 1 without a line between" signify?

Published on 01 Aug 2016 - 15:37

score 222 · Answer 7 · 2016-08-02 06:09:32

222

Views

Prove that $\displaystyle \sum_{1\leq k<j\leq n} \tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4} $

Published on 02 Aug 2016 - 6:09

score 3.4k · Answer 8 · 2016-08-03 15:55:21

3.4k

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Binomial Probability Problem

Published on 03 Aug 2016 - 15:55

score 685 · Answer 9 · 2026-03-31 04:20:43

685

Views

Proving Pascal's Identity algebraically, stuck with simplification

Published on 31 Mar 2026 - 4:20

score 196 · Answer 10 · 2026-04-10 19:08:25

196

Views

Is it possible a closed expression for $1^k + 2^k + ... + n^k$? If so, am I on the way?

Published on 10 Apr 2026 - 19:08

score 122 · Answer 11 · 2026-04-10 02:23:50

122

Views

Is there a combinatorial interpretation for the following sum?

Published on 10 Apr 2026 - 2:23

score 164 · Answer 12 · 2026-03-31 04:17:20

164

Views

Find the coefficient of the term $x^2$ in $\left(x+\frac 2x\right)^{4}$

Published on 31 Mar 2026 - 4:17

score 540 · Answer 13 · 2026-04-08 13:41:53

540

Views

Simplifying the product of multiple binomial expansions

Published on 08 Apr 2026 - 13:41

score 218 · Answer 14 · 2016-08-09 09:25:34

218

Views

Combinatorics proof of identity $\sum_{k=1}^{n}kC_n^k=n2^{n-1}$

Published on 09 Aug 2016 - 9:25

score 73 · Answer 15 · 2026-03-26 02:57:57

73

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How can I prove that $\left|\sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}\right| \leq \binom{n}{r}$?

Published on 26 Mar 2026 - 2:57

List Question

Prove that $\binom{2n}{n} > \prod_{p > n}^{p<2n}{p}$

Help on how to prove that $\ln{k\choose k/2}=\sum_{n=1}^{\infty}(-1)^{n-1}\left({k\over n}-\ln{n+k\over n}\right)$

Finding the coefficient of $x^r$ in an expansion.

Binomial identity: Clever idea for a short proof?

Generalization of the q-Binomial Theorem

What does "n over 1 without a line between" signify?

Prove that $\displaystyle \sum_{1\leq k<j\leq n} \tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4} $

Binomial Probability Problem

Proving Pascal's Identity algebraically, stuck with simplification

Is it possible a closed expression for $1^k + 2^k + ... + n^k$? If so, am I on the way?

Is there a combinatorial interpretation for the following sum?

Find the coefficient of the term $x^2$ in $\left(x+\frac 2x\right)^{4}$

Simplifying the product of multiple binomial expansions

Combinatorics proof of identity $\sum_{k=1}^{n}kC_n^k=n2^{n-1}$

How can I prove that $\left|\sum_{i=0}^r (-1)^i \binom{a}{i} \binom{n-a}{r-i}\right| \leq \binom{n}{r}$?

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