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

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Find the value of 1.1! + 2.2! + 3.3! + ......+n.n! (1) [#permalink] New post 09 Nov 2003, 05:15
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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, 16:30
choice 3 is right... is there any other way to solve... which short way you're talking about vicky??
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 [#permalink] New post 09 Nov 2003, 17:31
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, 18:22
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, 20:42
thats correct. It is the multiplication of numbers...
thanks
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 [#permalink] New post 11 Nov 2003, 00:38
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, 06:08
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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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, 06:08
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