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![[#8] 2 Min. Challenge : Factorial](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "[#8] 2 Min. Challenge : Factorial"){kind=icon}
Updated on: 09 Mar 2004, 12:17
Originally posted by Praetorian on 08 Mar 2004, 19:11.
Last edited by Praetorian on 09 Mar 2004, 12:17, edited 1 time in total.
Last edited by Praetorian on 09 Mar 2004, 12:17, edited 1 time in total.
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1. Time yourself
2. Solve this problem on a sheet of paper [ under 2 mins, if possible]
3. type in your solution and the time, if possible
What is the value of 1*1! + 2*2! + 3*3! + ......+n*n!
(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
(5) n! + 3
1. Time yourself
2. Solve this problem on a sheet of paper [ under 2 mins, if possible]
3. type in your solution and the time, if possible
What is the value of 1*1! + 2*2! + 3*3! + ......+n*n!
(1) n! +1
(2) (n+1)!
(3) (n+1)!-1
(4) (n+1)!+1
(5) n! + 3
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09 Mar 2004, 11:49
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praetorian123
again, i quote here.
Method 1
Consider n=2
1*1!+2*2!=1+4=5
check options
A=3 out
B=3!=6 out
C=6-1=5 OK
D=6+1=7 out
E=2+3=5 OK
consider n=3; S=5+18=23
C=23 OK
E=6+3=9 out
So, C.
Method 2
Let's see if we can manipulate all of the a*a! terms
a*a! = (a+1 - 1)*a! = (a+1)a! - a! = (a+1)! - a!
Hence, 1*1! + 2*2! + .... +n*n! = (2!-1!) + (3! - 2!) + .....(n + 1)! - n!.
If you add up all of the terms, you get: -1! + (n+1)! or (n+1)! - 1
which corresponds to answer (3).
