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Find a closed form for 
 \(S_n = 1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n!.\)

for integer n >= 1 Your response should have a factorial.

waffles  Feb 14, 2018
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2+0 Answers

 #1
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Here is the closed form, but now you have to prove it !!!

S(n) =[n + 1]! - 1

Hint: nn! = [n + 1]! - n!

Guest Feb 14, 2018
 #2
avatar+85673 
+1

 

 

I'm stealing something here that I learned from heureka.....so, really....he should get the credit  !!!!

 

He noted that

 

n * n!  =

( [ n + 1 ]  - 1) * n! =

(n + 1)n! - n!  =

(n + 1)! - n!

 

So  we have the following

 

     1 * 1!           =    2!  -  1!

+   2 * 2!           =    3!  -  2!

+   3 * 3!           =    4!  -  3!

+   4 * 4!           =    5!  -  4!

+   .......

+   n * n!           =  (n + 1)!  - (n)! 

 

The terms in red  will "cancel"  and we will  be left with

 

Sum  = ( n + 1)!  - 1!  =

 

Sum  =  (n + 1)!  -  1

 

 

cool cool cool

CPhill  Feb 14, 2018
edited by CPhill  Feb 14, 2018

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