combinations - Relation connecting $ (3n)!$, $3^n$ and $n ...
Jun 23, 2022 · Relation connecting (3n)!, 3 n and n! Ask Question Asked 3 years, 6 months ago Modified 3 years, 6 months ago
Simplifying fraction with factorials: $\\frac{(3(n+1))!}{(3n)!}$
Jan 23, 2017 · Now I realise using Sterling's formula would make everything easier, but my first approach was simplifying the factorial after applying the criterion I mentioned before. So, after …
What is the importance of the Collatz conjecture? [closed]
What delights me most about the Collatz conjecture is your observation about what the iteration does to the factorizations combined with an observation on the sizes of the numbers. …
Prove that $(3n)!/(3!)^n$ is an integer - Mathematics Stack Exchange
May 29, 2017 · Prove that (3n)! (3!)n (3 n)! (3!) n is an integer where n n is a non-negative integer. I can prove it with mathematical induction. Is there any other method?
Collatz conjecture but with $\ 3n-1\ $ instead of $\ 3n+1.\ $ Do …
Jun 28, 2021 · Because 3n+1 is the same as the absolute value of 3n-1 for negative numbers. So the question remains unanswered. If you found "lots" of answers that would be interesting, …
Prove by induction that $n^3 < 3^n$ - 3^n$ - Mathematics Stack …
Dec 22, 2016 · I dont know how I would find the root to the cubic without using numerical approximations or else but surely there is an easier way to show that the cubic $ 2n^3 -3n^2 …
Prove $3^n > n^2$ by induction - Mathematics Stack Exchange
Nov 10, 2015 · The hypothesis that 3n> n2 3 n> n 2 is used for the first inequality, and you can probably figure out what "sufficiently large n n " is for the last step. The 3n2> (n+1)2 3 n 2> (n …
Prove through induction that $3^n > n^3$ for $n \\geq 4$
I'm new to induction and have not done induction with inequalities before, so I get stuck at proving after the 3rd step. The question is: Use induction to show that ...
divisibility - Proving $n^3 + 3n^2 +2n$ is divisible by $6 ...
Jul 13, 2015 · The full question is: Factorise n3 + 3n2 + 2n n 3 + 3 n 2 + 2 n. Hence prove that when n n is a positive integer, n3 + 3n2 + 2n n 3 + 3 n 2 + 2 n is always divisible by 6 6. So i …
$3n+3$ conjecture - Mathematics Stack Exchange
May 23, 2023 · If you look at that remaining sequence starting with 3n+3 and ending in 3, and divide all its numbers by 3, then you get a normal Collatz sequence that starts with n+1 and …