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anonymous

  • 5 years ago

does anyone know how 2n+2! becomes (2n+2)(2n+1)

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  1. anonymous
    • 5 years ago
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    i guess it must b (2n +2)! ?

  2. anonymous
    • 5 years ago
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    yah

  3. anonymous
    • 5 years ago
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    n! = n(n-1)!

  4. anonymous
    • 5 years ago
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    (2n+2)(2n+1) (2n)....1

  5. anonymous
    • 5 years ago
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    (2n+2)!= (2n+2)(2n+1)!

  6. anonymous
    • 5 years ago
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    on cramster one of the problem shows it changes from (2n+2)! and becomes (2n+1)(2n+2) with no factorial sign behind it

  7. anonymous
    • 5 years ago
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    there is a facatorial behind it......

  8. anonymous
    • 5 years ago
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    i mean they don't show it like this (2n+2)(2n+1) (2n)....1

  9. anonymous
    • 5 years ago
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    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

  10. anonymous
    • 5 years ago
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    yah they dont show it decreasing it stops after 2n+1

  11. anonymous
    • 5 years ago
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    yes but there must be a factorial with it

  12. anonymous
    • 5 years ago
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    and there isnt

  13. anonymous
    • 5 years ago
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    As far as I know about factorials ...there should be...I could be wrong

  14. anonymous
    • 5 years ago
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    is there anyway you can look at the problem on cramster

  15. anonymous
    • 5 years ago
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    u can check on wolframalpha.com

  16. anonymous
    • 5 years ago
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    there must be factorial with it

  17. anonymous
    • 5 years ago
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    there has to be a factorial behind it, unless there is something else about your problem... cramster sometimes has mistakes

  18. anonymous
    • 5 years ago
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    the things is my teacher did the same thing so im confused

  19. anonymous
    • 5 years ago
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    the only way the factorial can drop out is if it cancelled out in a fraction

  20. anonymous
    • 5 years ago
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    then check did u post the right question?

  21. anonymous
    • 5 years ago
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    this is the actual problem determine of this series converges or diverges\[\sum_{n=1}^{\infty} (n!)^2/(2n) !\]

  22. anonymous
    • 5 years ago
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    use ratio test

  23. anonymous
    • 5 years ago
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    n u applied some test for the convergence of series?

  24. anonymous
    • 5 years ago
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    right

  25. anonymous
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    sigh i hate it when people post half the question... thigns cancel and become (n+1)^2/(2n+2)(2n+1)

  27. anonymous
    • 5 years ago
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    what is 2n!

  28. anonymous
    • 5 years ago
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    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

  29. anonymous
    • 5 years ago
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    yes i was able to catch my mistake thank you very much. i wanted to know how does 2n! decrease

  30. anonymous
    • 5 years ago
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    does it become (2n)(2n-1)(2n-2)..1

  31. anonymous
    • 5 years ago
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    yes

  32. anonymous
    • 5 years ago
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    hey thank u for ur help im gonna become a fan

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