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Serge Lang - Basic Mathematics - Sum of Binomial coefficients p.387 ex. 9

score 104 · Answer 1 · 2018-11-20 08:13:50

104

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Serge Lang - Basic Mathematics - Sum of Binomial coefficients p.387 ex. 9

Published on 20 Nov 2018 - 8:13

score 82 · Answer 2 · 2018-11-22 07:27:38

82

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General method to find sum of binomial

Published on 22 Nov 2018 - 7:27

score 152 · Answer 3 · 2018-11-24 02:57:18

152

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Does there exist a similar identity to $\binom{n}{k} = \binom{n}{n-k}$?

Published on 24 Nov 2018 - 2:57

score 287 · Answer 4 · 2018-11-25 13:11:20

287

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Showing $\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}$

Published on 25 Nov 2018 - 13:11

score 102 · Answer 5 · 2026-03-27 03:49:11

102

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How to show that $2^n > n$ without induction

Published on 27 Mar 2026 - 3:49

score 504 · Answer 6 · 2018-11-26 09:07:12

Given that the coefficients of the $6^{th}$ term and $16^{th}$ term in the Expansion of $(x + y) ^n$ are equal. Find the value of $n$.

score 544 · Answer 7 · 2026-03-26 05:55:55

544

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Factorial Proof - ${n \choose r-1}+{n \choose r}={n+1 \choose r}$

Published on 26 Mar 2026 - 5:55

score 39 · Answer 8 · 2018-11-28 17:13:57

39

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Is it possible to show

Published on 28 Nov 2018 - 17:13

score 539 · Answer 9 · 2026-03-25 23:52:18

539

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Is there a closed form for the sum of the cubes of the binomial coefficients?

Published on 25 Mar 2026 - 23:52

score 151 · Answer 10 · 2018-11-28 22:21:47

151

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Show there are $2^n$ forms a $n \times n$ matrix can take, preferably via Pascal's triangle

Published on 28 Nov 2018 - 22:21

score 307 · Answer 11 · 2026-03-25 06:21:41

307

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Weighted sum of product of binomial coefficients

Published on 25 Mar 2026 - 6:21

score 3.6k · Answer 12 · 2018-11-29 15:09:03

3.6k

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Counting paths through a checkers board (only moving diagonally)

Published on 29 Nov 2018 - 15:09

score 72 · Answer 13 · 2026-03-27 03:43:11

72

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One term of (2π+5)^n = 288000π^8, what's n?

Published on 27 Mar 2026 - 3:43

score 305 · Answer 14 · 2018-11-29 22:43:32

305

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Proof number-partitions $P_n = \sum_{k=1}^{n} P_{n,k}$ in positive summands

Published on 29 Nov 2018 - 22:43

score 142 · Answer 15 · 2026-03-27 20:21:06

142

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Derive $\sum_{s=r}^{\infty} \binom{m}{s} \binom{s}{r}(-1)^s=0 $ using an identity $(1 + x)^ m (1 + x)^{ -(r+1)} = (1 + x)^{ m-r-1}$

Published on 27 Mar 2026 - 20:21

List Question

Serge Lang - Basic Mathematics - Sum of Binomial coefficients p.387 ex. 9

General method to find sum of binomial

Does there exist a similar identity to $\binom{n}{k} = \binom{n}{n-k}$?

Showing $\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}$

How to show that $2^n > n$ without induction

Given that the coefficients of the $6^{th}$ term and $16^{th}$ term in the Expansion of $(x + y) ^n$ are equal. Find the value of $n$.

Factorial Proof - ${n \choose r-1}+{n \choose r}={n+1 \choose r}$

Is it possible to show

Is there a closed form for the sum of the cubes of the binomial coefficients?

Show there are $2^n$ forms a $n \times n$ matrix can take, preferably via Pascal's triangle

Weighted sum of product of binomial coefficients

Counting paths through a checkers board (only moving diagonally)

One term of (2π+5)^n = 288000π^8, what's n?

Proof number-partitions $P_n = \sum_{k=1}^{n} P_{n,k}$ in positive summands

Derive $\sum_{s=r}^{\infty} \binom{m}{s} \binom{s}{r}(-1)^s=0 $ using an identity $(1 + x)^ m (1 + x)^{ -(r+1)} = (1 + x)^{ m-r-1}$

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