List questions from binomial-theorem

Published on 04 Dec 2015 - 2:34

58

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Binomial Power Series Expansion

Published on 07 Dec 2015 - 10:40

481

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Powering a sum of two easily powered matrices

Published on 08 Dec 2015 - 22:04

42

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an intuition for $\sum {\frac{(n-2)!}{k_1!k_2!...k_n!}}=n^{n-2}$

Published on 31 Mar 2026 - 10:17

34

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How to fix the binomial formula for a specific application/bounds difference

Published on 10 Dec 2015 - 16:44

320

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Binomial Theorem- Finding the Greatest Term

Published on 10 Dec 2015 - 17:52

66

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Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$

Published on 10 Dec 2015 - 18:02

42

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Compute $\sum_{k=1}^{n+3} {(k+1)^2}\binom {n}{k}x^{k+1}$

Published on 10 Dec 2015 - 21:39

284

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Prove inequality using binomial theorem

Published on 11 Dec 2015 - 4:06

77

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Find $x^{10}y^6z^{-21}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem.

Published on 11 Dec 2015 - 20:39

5.3k

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Expanding $(a+b)^\frac{1}{2}$

Published on 12 Dec 2015 - 11:28

151

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Proving an inequality that looks related to the Binomial series,

Published on 14 Dec 2015 - 8:39

313

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Maximum remainder $(a-1)^n+(a+1)^n\mod a^2$ for $3\le a\le 1000$

Published on 14 Dec 2015 - 20:32

124

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Can be justified $\zeta(3)=\lim_{n\to\infty}-3\sum_{k=1}^n\sum_{\nu=0}^\infty\frac{(-1)^{\nu+1}}{\nu+1}\binom{3k^3-1}{\nu}$?

Published on 16 Dec 2015 - 16:21

38

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Compute $\sum_{k=-1}^{n=24}C(25,k+2)k2^k$

Published on 16 Dec 2015 - 16:33

List Question

Discrete math evaluate

Binomial Power Series Expansion

Powering a sum of two easily powered matrices

an intuition for $\sum {\frac{(n-2)!}{k_1!k_2!...k_n!}}=n^{n-2}$

How to fix the binomial formula for a specific application/bounds difference

Binomial Theorem- Finding the Greatest Term

Compute$\sum_{k=2}^{n+5}\frac {1}{k(k-1)}\binom {n+1}{k-2}2^k$

Compute $\sum_{k=1}^{n+3} {(k+1)^2}\binom {n}{k}x^{k+1}$

Prove inequality using binomial theorem

Find $x^{10}y^6z^{-21}$ in $(3x^2 -5y^{\frac{1}{2}} + z^{-3})^{24}$ with Binomial theorem.

Expanding $(a+b)^\frac{1}{2}$

Proving an inequality that looks related to the Binomial series,

Maximum remainder $(a-1)^n+(a+1)^n\mod a^2$ for $3\le a\le 1000$

Can be justified $\zeta(3)=\lim_{n\to\infty}-3\sum_{k=1}^n\sum_{\nu=0}^\infty\frac{(-1)^{\nu+1}}{\nu+1}\binom{3k^3-1}{\nu}$?

Compute $\sum_{k=-1}^{n=24}C(25,k+2)k2^k$

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