Question

A type of Combinatorial equality：$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$

score 120 · Answer 1 · 2020-04-12 04:10:39

120

Views

A type of Combinatorial equality：$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$

Published on 12 Apr 2020 - 4:10

score 243 · Answer 2 · 2020-04-13 04:24:14

243

Views

Identity of binomial coefficients: $\binom{n+1}{k+1}=\sum_{i=m}^{n+m-k}{\binom{n-i}{k-m}\binom{i}{m}}$

Published on 13 Apr 2020 - 4:24

score 240 · Answer 3 · 2026-03-25 12:50:49

240

Views

Let $x^2=y^2=1$ and $xy\neq yx$. There are $\binom{2n}{n}$ expressions of length $2n$ in $x$ and $y$ that are equal to $1$.

Published on 25 Mar 2026 - 12:50

score 186 · Answer 4 · 2026-03-25 14:25:19

186

Views

The number of pairs $(m,n)$ of coprime positive integers that divide $k$ is $d(k^2)$, where $d$ is the divisor counting function.

Published on 25 Mar 2026 - 14:25

score 106 · Answer 5 · 2020-04-18 15:24:37

106

Views

Stirling number of the first kind summation

Published on 18 Apr 2020 - 15:24

score 104 · Answer 6 · 2020-04-18 21:28:01

104

Views

I Need help creating an intuitive answer to the sum of $1(1!) + 2(2!) + 3(3!) +\cdots+ n(n!)$

Published on 18 Apr 2020 - 21:28

score 116 · Answer 7 · 2020-04-19 16:05:16

116

Views

A summation involving stirling numbers of the first kind

Published on 19 Apr 2020 - 16:05

score 98 · Answer 8 · 2020-04-24 14:23:33

98

Views

Prove combinatorically that $\sum_{i=1}^{n}2^{i-1}3^{n-i}=3^n-2^n$.

Published on 24 Apr 2020 - 14:23

score 319 · Answer 9 · 2020-04-24 17:16:57

319

Views

Binomial Coefficient Intuition

Published on 24 Apr 2020 - 17:16

score 655 · Answer 10 · 2020-04-26 06:02:54

655

Views

Use generating functions to prove that the number of partitions of a positive integer n into parts, each part at most 2, is [n/2] + 1

Published on 26 Apr 2020 - 6:02

score 74 · Answer 11 · 2020-04-28 22:04:32

74

Views

Can we simplify this summation involving factorials?

Published on 28 Apr 2020 - 22:04

score 108 · Answer 12 · 2020-04-29 14:30:09

108

Views

A Conceptual Intepretation of the identity

Published on 29 Apr 2020 - 14:30

score 1k · Answer 13 · 2020-04-29 17:55:08

1k

Views

Combinatorial proof that $\sum_0^n {n+k \choose n}{2n-k-1 \choose n-1} = {3n \choose n}$

Published on 29 Apr 2020 - 17:55

score 85 · Answer 14 · 2020-05-01 12:31:03

85

Views

Combinatory Explanation for $ \sum_{k=1}^{100} {{100} \choose {k}}2^k=3^{100}-1 $

Published on 01 May 2020 - 12:31

score 1.3k · Answer 15 · 2020-05-01 15:28:15

1.3k

Views

How to prove ${n+2 \choose 3}=1\cdot n + 2 \cdot (n - 1) + \ldots + n \cdot 1$?

Published on 01 May 2020 - 15:28

List Question

A type of Combinatorial equality：$\sum_{k=0}^{n}\binom{n}{k} \cos\frac{k}{2}\pi=2^{\frac{n}{2}}\cos\frac{n}{4}\pi.$

Identity of binomial coefficients: $\binom{n+1}{k+1}=\sum_{i=m}^{n+m-k}{\binom{n-i}{k-m}\binom{i}{m}}$

Let $x^2=y^2=1$ and $xy\neq yx$. There are $\binom{2n}{n}$ expressions of length $2n$ in $x$ and $y$ that are equal to $1$.

The number of pairs $(m,n)$ of coprime positive integers that divide $k$ is $d(k^2)$, where $d$ is the divisor counting function.

Stirling number of the first kind summation

I Need help creating an intuitive answer to the sum of $1(1!) + 2(2!) + 3(3!) +\cdots+ n(n!)$

A summation involving stirling numbers of the first kind

Prove combinatorically that $\sum_{i=1}^{n}2^{i-1}3^{n-i}=3^n-2^n$.

Binomial Coefficient Intuition

Use generating functions to prove that the number of partitions of a positive integer n into parts, each part at most 2, is [n/2] + 1

Can we simplify this summation involving factorials?

A Conceptual Intepretation of the identity

Combinatorial proof that $\sum_0^n {n+k \choose n}{2n-k-1 \choose n-1} = {3n \choose n}$

Combinatory Explanation for $ \sum_{k=1}^{100} {{100} \choose {k}}2^k=3^{100}-1 $

How to prove ${n+2 \choose 3}=1\cdot n + 2 \cdot (n - 1) + \ldots + n \cdot 1$?

Trending Questions

Popular # Hahtags

Popular Questions