Difficult Math Problem #97 - Algebra

[This topic has 8 member replies]
Forum Index GMAT Math Problem Solving
Free $100 Amazon.com Gift Card - Buy a GMAT course using a Beat The GMAT discount code between Mar 8-22 and get a $100 Amazon.com Gift Card. Learn more!
Post New Topic   Post Reply

800guy
GMAT Destroyer!

Default Avatar

Joined: 27 Jun 2006
Posts: 354

Thanks given: 1
Thanked 2 times in 2 posts
Topic: Difficult Math Problem #97 - Algebra
PostWed Feb 14, 2007 4:11 pm
Elapsed Time:
00:00
 Lap   Why a timer is critical to improving your score
Find 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) None of these


oa coming when some people answer/explain. from diff math doc.
Back to top
View user's profile Send private message
kandelaki
Rising GMAT Star

Default Avatar

Joined: 05 Feb 2007
Posts: 37

Thanks given: 0
Thanked 0 times in 0 posts
PostThu Feb 15, 2007 4:27 am
Sn = n ( 2a1 + (n - 1)d ) / 2

so Sn= n( 2*1.1!+(n-1)*1.1!)/2 =n(2*1.1!+1.1n!-1.1!)/2=n1.1!(2+n-1)/2=

=1.1n!(n+1)/2

i can not go any further...
Back to top
View user's profile Send private message
banona
Rising GMAT Star

Default Avatar

Joined: 30 Dec 2006
Posts: 38

Thanks given: 0
Thanked 0 times in 0 posts
PostThu Feb 15, 2007 7:39 am
Here is my attempt to solve this difficlt problem:

we are looking for the sum ( kK!) as K from 2 to n.
First , we can easily eliminate first and second choices, because the first one ( n!+1) is vritually too small to be the sum we are looking for;
The second choice ( n+1)! is even; and the sum we are looking for is odd;
Now choices are narrowed to the three choices that countain the expression ( n+1)!

Let's examine the following difference :
(n+1)! - [ 1*1! + 2*2! +........+ n *n! )] = (n+1)*n! - [ 1*1! + 2*2! +........+ n *n! )] = n! - [ 1*1! + 2*2! +........+ (n-1) *(n-1)! )] = n ( n-1)! - [ 1*1! + 2*2! +........+ (n-1) *(n-1)! )] = (n-1)(n-2)! - [ 1*1! + 2*2! +........+ (n-2) *(n-2)! )]
THE GENERAL FORMULA, when doing so (k) times, is
(n-K+1)(n-K)! - [ 1*1! + .......+ (n-k) *(n-K)! )]

Therfore at the end ; k is (n-1), so the sum becomes :
2*(1)! - [ 1*1!] = 1 = (n+1)! - [ 1*1! + 2*2! +........+ n *n! )] = 1
So,
[ 1*1! + 2*2! +........+ n *n! )] = (n+1)! -1


I hope I am not abusively simplifying things,
please, any comment !!!!!
Back to top
View user's profile Send private message
banona
Rising GMAT Star

Default Avatar

Joined: 30 Dec 2006
Posts: 38

Thanks given: 0
Thanked 0 times in 0 posts
PostThu Feb 15, 2007 8:06 am
Hy Kandelaki,
I wonder if you can use this general formula like :
Sn = n ( 2a1 + (n - 1)d ) / 2
I think it's only valid when adding consecutive numbers, like (1, 2,3,......n) or when numbers are equally far from each other; like consecutive evens ( 2; 4; 6; ......2n) or consecutives odds;
However, in our case 11! ; 22! ; 33!; ....... nn! are not consecutive numbers;



Can Mathematics tutors comment ?
Back to top
View user's profile Send private message
Mark Dabral
GMAT Instructor

Default Avatar

Joined: 26 Oct 2006
Posts: 43

Thanks given: 0
Thanked 2 times in 1 posts
Location: Berkeley, California
PostThu Feb 15, 2007 12:45 pm
hi guys,

i am sure you know that this question is really way out of GMAT league.

S = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ..... + (n-1)[(n-1)!] + n(n!)

S = [2-1](1!) + [3-1](2!) + [4-1](3!) + [5-1](4!) + ..... + (n-1)[(n-1)!] + [n+1 - 1](n!)

S = 2(1!) - 1! + 3(2!) - 2! + 4(3!) - 3! + 5(4!) - 4!+ ..... + n[(n-1)!] - (n-1)! + (n+1)(n!) - n!

S = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4!+ ..... + n! - (n-1)! + (n+1)! - n!

The terms 2!, 3!, 4!, and so on cancel out leaving only (n+1)! and the 1 term.

Therefore, S = (n+1)! - 1

Cheers,
Mark
_________________
Mark Dabral
GMAT Math Tutor
Berkeley, California
Back to top
View user's profile Send private message
800guy
GMAT Destroyer!

Default Avatar

Joined: 27 Jun 2006
Posts: 354

Thanks given: 1
Thanked 2 times in 2 posts
PostFri Feb 16, 2007 12:27 pm
oa:

1.1! + 2.2! + 3.3! + ......+n.n!
=1.1! + (3-1)2! + (4-1)3! +......+ ((n+1)-1) n!
=1.1!+3!-2!+4!-3!+.......+(n+1)!-n!

So it is (n+1)! -1 (Answer choice 4)
Back to top
View user's profile Send private message
gabriel
GMAT Destroyer!



Joined: 20 Dec 2006
Posts: 986

Thanks given: 138
Thanked 37 times in 34 posts
Location: India
PostSun Feb 18, 2007 5:31 am
Mark Dabral wrote:
hi guys,

i am sure you know that this question is really way out of GMAT league.

S = 1(1!) + 2(2!) + 3(3!) + 4(4!) + ..... + (n-1)[(n-1)!] + n(n!)

S = [2-1](1!) + [3-1](2!) + [4-1](3!) + [5-1](4!) + ..... + (n-1)[(n-1)!] + [n+1 - 1](n!)

S = 2(1!) - 1! + 3(2!) - 2! + 4(3!) - 3! + 5(4!) - 4!+ ..... + n[(n-1)!] - (n-1)! + (n+1)(n!) - n!

S = 2! - 1! + 3! - 2! + 4! - 3! + 5! - 4!+ ..... + n! - (n-1)! + (n+1)! - n!

The terms 2!, 3!, 4!, and so on cancel out leaving only (n+1)! and the 1 term.

Therefore, S = (n+1)! - 1

Cheers,
Mark
hi there, great effort..... but u know what such q are much more easier than they seem..... make use of the answer choices...


in this particular q all that has to be done is choose a value for n .... eg let
n=2 so the series will be 1*1!+ 2*2! = 5.... now substitute n=2 in the answer choices... and u will find that only ( n+1 ) ! - 1 will give u a value 5 for n=2..... hope that helps
Back to top
View user's profile Send private message
gabriel
GMAT Destroyer!



Joined: 20 Dec 2006
Posts: 986

Thanks given: 138
Thanked 37 times in 34 posts
Location: India
PostSun Feb 18, 2007 5:35 am
[quote="banona"]Hy Kandelaki,
I wonder if you can use this general formula like :
Sn = n ( 2a1 + (n - 1)d ) / 2
I think it's only valid when adding consecutive numbers, like (1, 2,3,......n) or when numbers are equally far from each other; like consecutive evens ( 2; 4; 6; ......2n) or consecutives odds;
However, in our case 11! ; 22! ; 33!; ....... nn! are not consecutive numbers;



Can Mathematics tutors comment ?[/quote

yup....u r rite.... that is the formula for a AP...ie for a series with equally spaced elements...the formula cant be used in this case
Back to top
View user's profile Send private message
Mark Dabral
GMAT Instructor

Default Avatar

Joined: 26 Oct 2006
Posts: 43

Thanks given: 0
Thanked 2 times in 1 posts
Location: Berkeley, California
PostSun Feb 18, 2007 3:42 pm
hi gabriel,

you are absolutely correct that if such a question would show up on the GMAT, then the fastest way to do it is to pick a small value for n and plug it in the answer choices. Typically, GMAC would not ask questions that can be easily figured out by picking a simple number, if it is possible to do so, then the question would be ranked as an easy one. This problem is of course not appropriate for the GMAT, but the key element to learn is to see the pattern and rearrange the terms. There is a significant emphasis on sequences beginning 2006, and many of the questions require one to rewrite the terms or see a recurring pattern, and do not require any specific knowledge of arithmetic and geometric series.

So if you do see a fairly complicated sequence with numbers on the GMAT, then look for a pattern that eliminates or merges multiple terms.

Cheers,
Mark
_________________
Mark Dabral
GMAT Math Tutor
Berkeley, California
Back to top
View user's profile Send private message
Display posts from previous:   

Post New Topic   Post Reply All times are GMT - 7 Hours
Page 1 of 1
 
Forum Index GMAT Math Problem Solving
Most Active Members in Last 30 Days
1. harsh.champ 461 posts
2. shashank.ism 366 posts
3. thephoenix 323 posts
4. kstv 320 posts
5. gmatmachoman 261 posts
See More Top Beat The GMAT Members
Most Active Experts in Last 30 Days
1. lunarpower
Manhattan GMAT Teacher 		87 posts
2. Stuart Kovinsky
Kaplan GMAT Teacher 		60 posts
3. Lisa Anderson
Stacy Blackman Consulting 		54 posts
4. Testluv
Kaplan GMAT Teacher 		49 posts
5. Bryant@VeritasPrep
Veritas Prep 		41 posts
See More Top Beat The GMAT Experts