Question

Prove $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers

score 50 · Answer 1 · 2018-12-12 23:25:01

50

Views

Prove $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers

Published on 12 Dec 2018 - 23:25

score 198 · Answer 2 · 2018-12-13 11:27:07

198

Views

Is it possibile to obtain the sum of the following series without using hypergeometric functions?

Published on 13 Dec 2018 - 11:27

score 121 · Answer 3 · 2018-12-15 09:43:33

121

Views

How find the limit of $\lim\limits_{n\to \infty }\left(n-1\right)!$

Published on 15 Dec 2018 - 9:43

score 197 · Answer 4 · 2018-12-15 18:08:53

197

Views

What would be the sequence of $n!!$ (double factorial)

Published on 15 Dec 2018 - 18:08

score 59 · Answer 5 · 2018-12-17 23:45:08

59

Views

Simplyfying factorials why is $(n+1)(n-1)!-(n-1)! = n(n-1)!$?

Published on 17 Dec 2018 - 23:45

score 81 · Answer 6 · 2018-12-19 09:22:27

81

Views

Turning product sequences into factorials

Published on 19 Dec 2018 - 9:22

score 288 · Answer 7 · 2018-12-22 12:58:59

288

Views

why is $\frac{k\cdot n!}{k!(n-k)!} = \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}$?

Published on 22 Dec 2018 - 12:58

score 58 · Answer 8 · 2018-12-22 15:26:00

58

Views

Further explanation for steps of an equation that proofs that $\sum^{n}_{k=0}k\cdot \binom{n}{k}=n\cdot2^{n-1}$

Published on 22 Dec 2018 - 15:26

score 55 · Answer 9 · 2018-12-24 00:48:22

55

Views

Need help understanding a step in an induction proof

Published on 24 Dec 2018 - 0:48

score 155 · Answer 10 · 2018-12-24 23:30:12

155

Views

Calculate $\sum_{k=l}^{2n} \binom{2n+k}{2k} \frac{(2k-1)!!}{(k-l)!} (-1)^k$

Published on 24 Dec 2018 - 23:30

score 331 · Answer 11 · 2018-12-26 17:42:06

331

Views

Probability an ace lies behind first ace

Published on 26 Dec 2018 - 17:42

score 181 · Answer 12 · 2026-03-25 06:09:33

181

Views

Twin Prime Formula

Published on 25 Mar 2026 - 6:09

score 547 · Answer 13 · 2019-01-02 21:30:57

547

Views

Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial

Published on 02 Jan 2019 - 21:30

score 1k · Answer 14 · 2019-01-03 12:54:44

1k

Views

Show that $\frac{(2n)!!}{(2n+1)!!}$ converges.

Published on 03 Jan 2019 - 12:54

score 62 · Answer 15 · 2019-01-06 20:54:29

62

Views

A question on the approximation of $\ln(x!)$ for small x

Published on 06 Jan 2019 - 20:54

List Question

Prove $2\times1!+5\times2!+10\times3!+...+(n^2+1)n!=n(n+1)!$ for all positive integers

Is it possibile to obtain the sum of the following series without using hypergeometric functions?

How find the limit of $\lim\limits_{n\to \infty }\left(n-1\right)!$

What would be the sequence of $n!!$ (double factorial)

Simplyfying factorials why is $(n+1)(n-1)!-(n-1)! = n(n-1)!$?

Turning product sequences into factorials

why is $\frac{k\cdot n!}{k!(n-k)!} = \frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}$?

Further explanation for steps of an equation that proofs that $\sum^{n}_{k=0}k\cdot \binom{n}{k}=n\cdot2^{n-1}$

Need help understanding a step in an induction proof

Calculate $\sum_{k=l}^{2n} \binom{2n+k}{2k} \frac{(2k-1)!!}{(k-l)!} (-1)^k$

Probability an ace lies behind first ace

Twin Prime Formula

Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial

Show that $\frac{(2n)!!}{(2n+1)!!}$ converges.

A question on the approximation of $\ln(x!)$ for small x

Trending Questions

Popular # Hahtags

Popular Questions