anonymous
  • anonymous
does anyone know how 2n+2! becomes (2n+2)(2n+1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i guess it must b (2n +2)! ?
anonymous
  • anonymous
yah
anonymous
  • anonymous
n! = n(n-1)!

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anonymous
  • anonymous
(2n+2)(2n+1) (2n)....1
anonymous
  • anonymous
(2n+2)!= (2n+2)(2n+1)!
anonymous
  • anonymous
on cramster one of the problem shows it changes from (2n+2)! and becomes (2n+1)(2n+2) with no factorial sign behind it
anonymous
  • anonymous
there is a facatorial behind it......
anonymous
  • anonymous
i mean they don't show it like this (2n+2)(2n+1) (2n)....1
anonymous
  • anonymous
I meant that u keep decreasing by 1 until u reach 1 another way of writing it (2n+2)(2n+1) (2n)...3.2.1
anonymous
  • anonymous
yah they dont show it decreasing it stops after 2n+1
anonymous
  • anonymous
yes but there must be a factorial with it
anonymous
  • anonymous
and there isnt
anonymous
  • anonymous
As far as I know about factorials ...there should be...I could be wrong
anonymous
  • anonymous
is there anyway you can look at the problem on cramster
anonymous
  • anonymous
u can check on wolframalpha.com
anonymous
  • anonymous
there must be factorial with it
anonymous
  • anonymous
there has to be a factorial behind it, unless there is something else about your problem... cramster sometimes has mistakes
anonymous
  • anonymous
the things is my teacher did the same thing so im confused
anonymous
  • anonymous
the only way the factorial can drop out is if it cancelled out in a fraction
anonymous
  • anonymous
then check did u post the right question?
anonymous
  • anonymous
this is the actual problem determine of this series converges or diverges\[\sum_{n=1}^{\infty} (n!)^2/(2n) !\]
anonymous
  • anonymous
use ratio test
anonymous
  • anonymous
n u applied some test for the convergence of series?
anonymous
  • anonymous
right
anonymous
  • anonymous
i did but i didnt correct answer if someone can please show me how to do this it can help me with the rest of my problems
anonymous
  • anonymous
sigh i hate it when people post half the question... thigns cancel and become (n+1)^2/(2n+2)(2n+1)
anonymous
  • anonymous
what is 2n!
anonymous
  • anonymous
yes it converges, so you use ratio test... an+1/an= [ ((n+1)!)^2/(2(n+1))! ]/ [ (n!)^2/(2n)! ] flipping fractions to get 1 whole one.. (2n)!(n+1)!(n+1)!/(n!)(n!)(2n+2)! which is where your question arises, like I said, the only reason the factorial disappeared is cause it got cancelled, which you always want to do when doing ratio test, so you simply keep pulling your factorials apart until they can cancel so on top (n+1)!=(n+1)(n!), which you do twice to get rid of both the n! on the bottom then on the bottom you have (2n+2)! so you pull the factorial apart twice to get (2n+2)(2n+1)(2n)! then the (2n)! can cancel with the one on top and you get left with (n+1)^2/(2n+2)(2n+1). take the limit and it should be 1/4 so it converges
anonymous
  • anonymous
yes i was able to catch my mistake thank you very much. i wanted to know how does 2n! decrease
anonymous
  • anonymous
does it become (2n)(2n-1)(2n-2)..1
anonymous
  • anonymous
yes
anonymous
  • anonymous
hey thank u for ur help im gonna become a fan

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