Question

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

Answer 1

Shouldn't the second to last term really be 2013!/2012 ?

Answer 2

Oopsss..sorry i was typo there. Yeah the second last should be 2013!/2012

Answer 3

Fun problem I'll say that! I don't know where to begin hmm...

Answer 4

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

Answer 5

So our sequence is \(a_n= (-1)^{n+1}\dfrac{(n+2)!}{n+1}\)

Answer 6

I give up!! hehehe.. it's above my head!!

Answer 7

What you deleted everything AND gave up?!

Answer 8

Damn it

Answer 9

@Loser66 http://www.brainyquote.com/quotes/quotes/a/alberteins109012.html

Answer 10

hey! what do you need help with?

Answer 11

need a way and solution :)

Answer 12

Isn't that \[(-1)^{n+1} (n!+(n+1)!)\]

Answer 13

Telescopes quite nicely

Answer 14

right :)

Answer 15

So what's next @mukushla ?

Answer 16

do you see see how \(\dfrac{(n+2)!}{n+1} \) simplifies to \(n! + (n+1)!\) ?

Answer 17

(n+1)!/n
= (n+1)n(n-1)!/n
= (n+1)(n-1)!
hmmm...

Answer 18

\[\begin{align}\dfrac{(n+2)!}{n+1} &= \dfrac{(n+2)(n+1)n!}{n+1} \\~\\
&= (n+2)n! = (\color{blue}{n+1}+1)n!\\~\\
& = (n+1)n! + n! \\~\\
&= (n+1)!+n!\end{align}\]

Answer 19

ah yes i see now :) sorry i look n+1 but should be n+2. Whats next ?

Answer 20

familiar with sigma notation ? \(\sum\)

Answer 21

the given sum is same as :
\[ [(1+1)!+1!] - [(2+1)!+2!] + [(3+1)!+3!] -\cdots -[(2012+1)!+2012!] \]

Answer 22

Wooooooooooooooah!! it is nice.

Answer 23

or

\[ [2!+1!] - [3!+2!] + [4!+3!] -\cdots -[2013!+2012!] \]

Answer 24

please medal loser/mukushla, not meh

Answer 25

You work, why medal me?

Answer 26

Looks the series is nice but not sure which numbers can be cancels ?

Answer 27

i didnt use my brain, i just used ur work for general term and mukushla's idea of telescoping