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Find the value of 1.1! + 2.2! + 3.3! + ......+n.n! (1)

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Senior Manager
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Find the value of 1.1! + 2.2! + 3.3! + ......+n.n! (1) [#permalink] New post 09 Nov 2003, 04:15
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

83% (01:04) correct 17% (00:00) wrong based on 2 sessions
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) n! + 3

hint: one extremely short and simple way
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 [#permalink] New post 09 Nov 2003, 15:30
choice 3 is right... is there any other way to solve... which short way you're talking about vicky??
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shubhangi

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 [#permalink] New post 09 Nov 2003, 16:31
How exactly do you calculate factorial of decimals? Isn't factorial only calculated from integers?!?
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 [#permalink] New post 09 Nov 2003, 17:22
OOOPS. I did not even think that it was decimal. I thought it was multiplication. I thought the question in the following way.

1*1! + 2*2! + 3*3! + ....... + n*n!

Vicki,

Is my interpretation of the question correct? Please let us know.

Thanks
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 [#permalink] New post 09 Nov 2003, 19:42
thats correct. It is the multiplication of numbers...
thanks
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 [#permalink] New post 10 Nov 2003, 23:38
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.

Any other approach.
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Re: PS: n! [#permalink] New post 20 Nov 2003, 05:08
Vicky wrote:
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) n! + 3

hint: one extremely short and simple way


Had to think about this one:

You can solve this easily using substitution, but that's no fun.

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).
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Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Re: PS: n!   [#permalink] 20 Nov 2003, 05:08
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