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anonymous

  • one year ago

Evaluate : 3!/2 - 4!/3 + 5!/4 - 6!/5 + .... + 2013!/2014 - 2014!/2013

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  1. Empty
    • one year ago
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    Shouldn't the second to last term really be 2013!/2012 ?

  2. anonymous
    • one year ago
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    Oopsss..sorry i was typo there. Yeah the second last should be 2013!/2012

  3. Empty
    • one year ago
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    Fun problem I'll say that! I don't know where to begin hmm...

  4. Loser66
    • one year ago
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    Another approaching \(a_1= \dfrac{3!}{2}\\a_3=\dfrac{5!}{4}\\a_5=\dfrac{7!}{6}\\--------\) while \(a_2= -\dfrac{4!}{3}\\a_4=-\dfrac{6!}{5}\\a_6=-\dfrac{8!}{7}---------\)

  5. Loser66
    • one year ago
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    So our sequence is \(a_n= (-1)^{n+1}\dfrac{(n+2)!}{n+1}\)

  6. Loser66
    • one year ago
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    I give up!! hehehe.. it's above my head!!

  7. Empty
    • one year ago
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    What you deleted everything AND gave up?!

  8. Empty
    • one year ago
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    Damn it

  9. Empty
    • one year ago
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    @Loser66 http://www.brainyquote.com/quotes/quotes/a/alberteins109012.html

  10. anonymous
    • one year ago
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    hey! what do you need help with?

  11. anonymous
    • one year ago
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    need a way and solution :)

  12. anonymous
    • one year ago
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    Isn't that \[(-1)^{n+1} (n!+(n+1)!)\]

  13. anonymous
    • one year ago
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    Telescopes quite nicely

  14. anonymous
    • one year ago
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    right :)

  15. anonymous
    • one year ago
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    So what's next @mukushla ?

  16. ganeshie8
    • one year ago
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    do you see see how \(\dfrac{(n+2)!}{n+1} \) simplifies to \(n! + (n+1)!\) ?

  17. anonymous
    • one year ago
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    (n+1)!/n = (n+1)n(n-1)!/n = (n+1)(n-1)! hmmm...

  18. ganeshie8
    • one year ago
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    \[\begin{align}\dfrac{(n+2)!}{n+1} &= \dfrac{(n+2)(n+1)n!}{n+1} \\~\\ &= (n+2)n! = (\color{blue}{n+1}+1)n!\\~\\ & = (n+1)n! + n! \\~\\ &= (n+1)!+n!\end{align}\]

  19. anonymous
    • one year ago
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    ah yes i see now :) sorry i look n+1 but should be n+2. Whats next ?

  20. ganeshie8
    • one year ago
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    familiar with sigma notation ? \(\sum\)

  21. ganeshie8
    • one year ago
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    the given sum is same as : \[ [(1+1)!+1!] - [(2+1)!+2!] + [(3+1)!+3!] -\cdots -[(2012+1)!+2012!] \]

  22. Loser66
    • one year ago
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    Wooooooooooooooah!! it is nice.

  23. ganeshie8
    • one year ago
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    or \[ [2!+1!] - [3!+2!] + [4!+3!] -\cdots -[2013!+2012!] \]

  24. ganeshie8
    • one year ago
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    please medal loser/mukushla, not meh

  25. Loser66
    • one year ago
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    You work, why medal me?

  26. anonymous
    • one year ago
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    Looks the series is nice but not sure which numbers can be cancels ?

  27. ganeshie8
    • one year ago
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    i didnt use my brain, i just used ur work for general term and mukushla's idea of telescoping

  28. Loser66
    • one year ago
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    |dw:1439641996688:dw|

  29. anonymous
    • one year ago
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    1! - 2012! as finally ?

  30. ganeshie8
    • one year ago
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    \[ [2!+1!] - [3!+2!] + [4!+3!] -\cdots -[2013!+2012!] \] Let me break it like below : \[ \begin{align} &[2!+1!]\\~\\ & - [3!+2!] \\~\\ &+ [4!+3!] \\~\\ &-\cdots \\~\\ &-[2013!+2012!] \end{align}\]

  31. ganeshie8
    • one year ago
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    1! - 2012! is wrong, try again

  32. anonymous
    • one year ago
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    Ah looks diagonal numbers can be cancels. The rest is 1! - 2013! Hehe, thanks @ganeshie8 for ur leader

  33. ganeshie8
    • one year ago
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    np :)

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