Question

Expression for the sum $S = \sum_{i = 0}^{n - 1} {2n \choose i} \cdot x^i \cdot (1 - x)^{2n - i}$

score 81 · Answer 1 · 2019-05-20 21:34:34

81

Views

Expression for the sum $S = \sum_{i = 0}^{n - 1} {2n \choose i} \cdot x^i \cdot (1 - x)^{2n - i}$

Published on 20 May 2019 - 21:34

score 47 · Answer 2 · 2019-05-21 13:48:51

47

Views

Minimize $E(m)=\sum_{i = 0}^{n-1} {m \choose i} \cdot [c \cdot (\frac{N}{m})^2]^i \cdot[1-c \cdot (\frac{N}{m})^2]^{m-i}$

Published on 21 May 2019 - 13:48

score 142 · Answer 3 · 2019-05-21 15:10:49

142

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How to proof the Pascal's identity using the identity $(a+b)^n=(a+b)^{n-1}a+(a+b)^{n-1}b$?

Published on 21 May 2019 - 15:10

score 68 · Answer 4 · 2019-05-22 17:22:45

68

Views

What's the underlying combinatoric idea behind $(k-1)!\binom{n}{k}=\sum_{i=1}^{n-k+1}\frac{\left(n-i\right)!}{(n-k+1-i)!}$?

Published on 22 May 2019 - 17:22

score 83 · Answer 5 · 2026-03-26 22:15:25

83

Views

How is $\binom{2n}{n} \sum_{k=0}^{n}\binom{n}{k} \binom{n}{n-k}=(\binom{2n}{n})^{2}$

Published on 26 Mar 2026 - 22:15

score 286 · Answer 6 · 2019-05-23 02:50:14

286

Views

Prove $\sum_{k=j}^{n}{k \choose j}{n \choose k}{n-j \choose k+j}={2j \choose j}{2(n-j)\choose n-j}{n-j \choose 2j}{n+j \choose j}^{-1}$

Published on 23 May 2019 - 2:50

score 293 · Answer 7 · 2026-03-25 07:53:26

293

Views

The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is?

Published on 25 Mar 2026 - 7:53

score 208 · Answer 8 · 2019-05-23 17:10:53

208

Views

Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$

Published on 23 May 2019 - 17:10

score 78 · Answer 9 · 2019-05-23 20:52:06

78

Views

Summation with 3 binomial coefficients

Published on 23 May 2019 - 20:52

score 151 · Answer 10 · 2019-05-24 14:51:05

151

Views

Prove a sum of sums equals n choose k

Published on 24 May 2019 - 14:51

score 55 · Answer 11 · 2019-05-24 21:30:27

55

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How to find this kind of relationship: $x-\frac{1}{x}=A$ and $x+\frac{1}{x}=\sqrt{A^2+4}$?

Published on 24 May 2019 - 21:30

score 194 · Answer 12 · 2019-05-26 00:39:29

194

Views

Simplifying (and/or bounding) a sum of product of binomial coefficients

Published on 26 May 2019 - 0:39

score 179 · Answer 13 · 2026-03-26 01:34:38

179

Views

Prove identity $\sum^n_{k = 0} \binom {r+k} {r} = \binom{r+n+1} {r+1}$ using lattice paths

Published on 26 Mar 2026 - 1:34

score 50 · Answer 14 · 2019-05-27 14:09:59

50

Views

How to compute $ \sum_{2k \le n} (-3)^k \binom{n}{2k} $

Published on 27 May 2019 - 14:09

score 60 · Answer 15 · 2019-05-27 20:58:02

60

Views

Calculating integral $\int_{0}^{1}x^{k+1}\cdot (1-x)^l\cdot dx$

Published on 27 May 2019 - 20:58

List Question

Expression for the sum $S = \sum_{i = 0}^{n - 1} {2n \choose i} \cdot x^i \cdot (1 - x)^{2n - i}$

Minimize $E(m)=\sum_{i = 0}^{n-1} {m \choose i} \cdot [c \cdot (\frac{N}{m})^2]^i \cdot[1-c \cdot (\frac{N}{m})^2]^{m-i}$

How to proof the Pascal's identity using the identity $(a+b)^n=(a+b)^{n-1}a+(a+b)^{n-1}b$?

What's the underlying combinatoric idea behind $(k-1)!\binom{n}{k}=\sum_{i=1}^{n-k+1}\frac{\left(n-i\right)!}{(n-k+1-i)!}$?

How is $\binom{2n}{n} \sum_{k=0}^{n}\binom{n}{k} \binom{n}{n-k}=(\binom{2n}{n})^{2}$

Prove $\sum_{k=j}^{n}{k \choose j}{n \choose k}{n-j \choose k+j}={2j \choose j}{2(n-j)\choose n-j}{n-j \choose 2j}{n+j \choose j}^{-1}$

The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is?

Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$

Summation with 3 binomial coefficients

Prove a sum of sums equals n choose k

How to find this kind of relationship: $x-\frac{1}{x}=A$ and $x+\frac{1}{x}=\sqrt{A^2+4}$?

Simplifying (and/or bounding) a sum of product of binomial coefficients

Prove identity $\sum^n_{k = 0} \binom {r+k} {r} = \binom{r+n+1} {r+1}$ using lattice paths

How to compute $ \sum_{2k \le n} (-3)^k \binom{n}{2k} $

Calculating integral $\int_{0}^{1}x^{k+1}\cdot (1-x)^l\cdot dx$

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