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Difficult Math Problem #97 - Algebra

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by 800guy » Wed Feb 14, 2007 3:11 pm
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.
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by kandelaki » Thu Feb 15, 2007 3:27 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...
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by banona » Thu Feb 15, 2007 6:39 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 !!!!!
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by banona » Thu Feb 15, 2007 7:06 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 ?
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by 800guy » Fri Feb 16, 2007 11:27 am
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)
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Re: Another approach

by gabriel » Sun Feb 18, 2007 4:31 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
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by gabriel » Sun Feb 18, 2007 4:35 am
banona wrote: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
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