anonymous
  • anonymous
Evaluate : 3!/2 - 4!/3 + 5!/4 - 6!/5 + .... + 2013!/2014 - 2014!/2013
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Empty
  • Empty
Shouldn't the second to last term really be 2013!/2012 ?
anonymous
  • anonymous
Oopsss..sorry i was typo there. Yeah the second last should be 2013!/2012
Empty
  • Empty
Fun problem I'll say that! I don't know where to begin hmm...

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More answers

Loser66
  • Loser66
Another approaching \(a_1= \dfrac{3!}{2}\\a_3=\dfrac{5!}{4}\\a_5=\dfrac{7!}{6}\\--------\) while \(a_2= -\dfrac{4!}{3}\\a_4=-\dfrac{6!}{5}\\a_6=-\dfrac{8!}{7}---------\)
Loser66
  • Loser66
So our sequence is \(a_n= (-1)^{n+1}\dfrac{(n+2)!}{n+1}\)
Loser66
  • Loser66
I give up!! hehehe.. it's above my head!!
Empty
  • Empty
What you deleted everything AND gave up?!
Empty
  • Empty
Damn it
Empty
  • Empty
@Loser66 http://www.brainyquote.com/quotes/quotes/a/alberteins109012.html
anonymous
  • anonymous
hey! what do you need help with?
anonymous
  • anonymous
need a way and solution :)
anonymous
  • anonymous
Isn't that \[(-1)^{n+1} (n!+(n+1)!)\]
anonymous
  • anonymous
Telescopes quite nicely
anonymous
  • anonymous
right :)
anonymous
  • anonymous
So what's next @mukushla ?
ganeshie8
  • ganeshie8
do you see see how \(\dfrac{(n+2)!}{n+1} \) simplifies to \(n! + (n+1)!\) ?
anonymous
  • anonymous
(n+1)!/n = (n+1)n(n-1)!/n = (n+1)(n-1)! hmmm...
ganeshie8
  • ganeshie8
\[\begin{align}\dfrac{(n+2)!}{n+1} &= \dfrac{(n+2)(n+1)n!}{n+1} \\~\\ &= (n+2)n! = (\color{blue}{n+1}+1)n!\\~\\ & = (n+1)n! + n! \\~\\ &= (n+1)!+n!\end{align}\]
anonymous
  • anonymous
ah yes i see now :) sorry i look n+1 but should be n+2. Whats next ?
ganeshie8
  • ganeshie8
familiar with sigma notation ? \(\sum\)
ganeshie8
  • ganeshie8
the given sum is same as : \[ [(1+1)!+1!] - [(2+1)!+2!] + [(3+1)!+3!] -\cdots -[(2012+1)!+2012!] \]
Loser66
  • Loser66
Wooooooooooooooah!! it is nice.
ganeshie8
  • ganeshie8
or \[ [2!+1!] - [3!+2!] + [4!+3!] -\cdots -[2013!+2012!] \]
ganeshie8
  • ganeshie8
please medal loser/mukushla, not meh
Loser66
  • Loser66
You work, why medal me?
anonymous
  • anonymous
Looks the series is nice but not sure which numbers can be cancels ?
ganeshie8
  • ganeshie8
i didnt use my brain, i just used ur work for general term and mukushla's idea of telescoping
Loser66
  • Loser66
|dw:1439641996688:dw|
anonymous
  • anonymous
1! - 2012! as finally ?
ganeshie8
  • ganeshie8
\[ [2!+1!] - [3!+2!] + [4!+3!] -\cdots -[2013!+2012!] \] Let me break it like below : \[ \begin{align} &[2!+1!]\\~\\ & - [3!+2!] \\~\\ &+ [4!+3!] \\~\\ &-\cdots \\~\\ &-[2013!+2012!] \end{align}\]
ganeshie8
  • ganeshie8
1! - 2012! is wrong, try again
anonymous
  • anonymous
Ah looks diagonal numbers can be cancels. The rest is 1! - 2013! Hehe, thanks @ganeshie8 for ur leader
ganeshie8
  • ganeshie8
np :)

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