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Anonymous Fri 27 Dec 15:53:07 2019 No.11257326 View Reply Original Report

Quoted By: >>11257521 >>11257536 >>11257539 >>11257615 >>11257976 >>11258020 >>11258035 >>11258270

How to get right from left? I'm bad at factorials so help me please.

Anonymous

Anonymous Fri 27 Dec 2019 16:04:15 No.11257348 Report

Quoted By: >>11258035

If n = 1 then

3!^2 = 2! x 4!

36 = 48 ?

Anonymous

Anonymous Fri 27 Dec 2019 17:06:22 No.11257456 Report

Quoted By:

what I can get you is
$((n+2)!)^{n+1} = ((n+1)!\cdot (n+2))^n\cdot (n+2)! = ((n+1)!)^n\cdot (n+2)! \cdot (n+2)^n$

I don't think
$(n+2)^n = (2n+2)! / (n+2)!$

Anonymous

Anonymous Fri 27 Dec 2019 17:47:37 No.11257521 Report

Quoted By:

>>11257326
I dont think thats true.
Try proving by induction first.

Anonymous

Anonymous Fri 27 Dec 2019 17:55:34 No.11257536 Report

Quoted By:

>>11257326
You can't.
Just suppose 2n+1 is prime in the (2n+2)! on the right.

Anonymous

Anonymous Fri 27 Dec 2019 17:57:16 No.11257539 Report

Quoted By:

>>11257326
it's not true because $(2n+2)! \neq (n+1)!(n+2)^{n+1}$ (I hope it's obvious)

Anonymous

Anonymous Fri 27 Dec 2019 18:34:46 No.11257615 Report

Quoted By:

>>11257326
It is not equal.

Anonymous

Anonymous Fri 27 Dec 2019 22:13:00 No.11257976 Report

Quoted By:

>>11257326
bruh, just manipulate the left side and you will see clearly why the equality is not true

Anonymous

Anonymous Fri 27 Dec 2019 22:45:51 No.11258020 Report

Quoted By:

>>11257326
$<br />((n+2)!)^{n+1}=((n+1)!)^n\cdot(2n+2)!\\<br />(n+2)^{n}=\frac{(2n+2)!}{(n+2)!}=\prod_{k=n+3}^{2n+2} k<br />$
Bullshit.

Anonymous

Anonymous Fri 27 Dec 2019 22:55:42 No.11258035 Report

Quoted By:

>>11257326
this >>11257348

((n+2)!)^(n+1) = (3!)^2 = 36

((n+1)!)^n = 2
(2n+2)! = 4! = 24

[((n+1)!)^n]*[(2n+2)!] = 24*2 = 48 != 36

Anonymous

Anonymous Sat 28 Dec 2019 00:52:28 No.11258270 Report

Quoted By:

>>11257326
Maybe n is like 12 or something. Gotta be something after all.

Anonymous

Anonymous Sat 28 Dec 2019 00:52:42 No.11258271 Report

Quoted By:

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