Question

Proving $\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$

score 40 · Answer 1 · 2020-03-17 05:17:03

40

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Proving $\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$

Published on 17 Mar 2020 - 5:17

score 5.1k · Answer 2 · 2026-03-30 04:25:31

5.1k

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Binomial expansion of $(1+x)^{2n}$ and proof that $\binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$

Published on 30 Mar 2026 - 4:25

score 61 · Answer 3 · 2026-03-30 04:23:49

61

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Combinatory Analisis and Newton's Binomial

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score 65 · Answer 4 · 2020-03-20 11:00:34

65

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The simplest form for $x_j$ if $\sum_{j=1}^n \frac{x_j}{i+j-1}=1,~ 1\le i \le n$

Published on 20 Mar 2020 - 11:00

score 108 · Answer 5 · 2020-03-21 10:01:34

108

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Binomial coefficient when the order is smallest to largest

Published on 21 Mar 2020 - 10:01

score 28 · Answer 6 · 2026-03-25 09:25:44

28

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This Summation yields different results when a specific identity from Combinatorics is used.

Published on 25 Mar 2026 - 9:25

score 164 · Answer 7 · 2020-03-22 14:59:02

164

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Binomial series problem $\lim_{n\to\infty}\left[\sum_{i,j}{2i\choose i}{2j\choose j}\right]^{1/n}=4$

Published on 22 Mar 2020 - 14:59

score 59 · Answer 8 · 2020-03-22 15:32:52

59

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$\binom{2.10^n}{10^n}$ is divisible by 10 but not by 100

Published on 22 Mar 2020 - 15:32

score 95 · Answer 9 · 2020-03-23 01:11:20

95

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Evaluating $\sum_{0\leq k \leq l \leq n}\binom{k}{2}\binom{l}{k}\binom{n}{l}$

Published on 23 Mar 2020 - 1:11

score 180 · Answer 10 · 2020-03-23 02:16:51

180

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How to prove $f(n)=\sum_{k=0}^{n}C_{2n+k}^{n}C_{n-1+k}^{n-1}\equiv1\pmod2$?

Published on 23 Mar 2020 - 2:16

score 54 · Answer 11 · 2026-03-30 06:12:42

54

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Prove that $\sum_{i=1}^n \sqrt{ \binom{n}{i}} \le [n(2^n -1) ]^{1/2}$ for $n \ge 2$

Published on 30 Mar 2026 - 6:12

score 59 · Answer 12 · 2020-03-23 14:38:38

Show that for $n\geqslant1,\ \displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}$

score 45 · Answer 13 · 2020-03-23 17:07:01

45

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A proof for sum including binomial

Published on 23 Mar 2020 - 17:07

score 58 · Answer 14 · 2026-04-06 21:00:17

58

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Proof verification : A weird identity involving sums

Published on 06 Apr 2026 - 21:00

score 43 · Answer 15 · 2026-03-28 00:46:44

43

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Calculating a Binomial Expression

Published on 28 Mar 2026 - 0:46

List Question

Proving $\sum_{i=0}^{p}{\sum_{j=0}^{q+i}{\sum_{k=0}^{r+j}{\binom{p}{i}\binom{q+i}{j}\binom{r+j}{k}}}}=4^{p}3^{q}2^{r}$

Binomial expansion of $(1+x)^{2n}$ and proof that $\binom{2n}{n} = \binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$

Combinatory Analisis and Newton's Binomial

The simplest form for $x_j$ if $\sum_{j=1}^n \frac{x_j}{i+j-1}=1,~ 1\le i \le n$

Binomial coefficient when the order is smallest to largest

This Summation yields different results when a specific identity from Combinatorics is used.

Binomial series problem $\lim_{n\to\infty}\left[\sum_{i,j}{2i\choose i}{2j\choose j}\right]^{1/n}=4$

$\binom{2.10^n}{10^n}$ is divisible by 10 but not by 100

Evaluating $\sum_{0\leq k \leq l \leq n}\binom{k}{2}\binom{l}{k}\binom{n}{l}$

How to prove $f(n)=\sum_{k=0}^{n}C_{2n+k}^{n}C_{n-1+k}^{n-1}\equiv1\pmod2$?

Prove that $\sum_{i=1}^n \sqrt{ \binom{n}{i}} \le [n(2^n -1) ]^{1/2}$ for $n \ge 2$

Show that for $n\geqslant1,\ \displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}$

A proof for sum including binomial

Proof verification : A weird identity involving sums

Calculating a Binomial Expression

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