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prove that \(\displaystyle\sum_{r=1}^{n}r! \)   is divisible by 3. (n should be greater than or equal to 2)

rosala  Feb 3, 2018
 #1
avatar+86859 
+1

Notice, rosala, that if n= 2, we have...

 

1! + 2!  =  1 + 2  =  3   which is, of course, divisible by 3

 

If n   ≥ 3  we can write

 

1!  +  2!  +  3!   +  4! +   .........+ (n - 2)! + (n - 1)!  +  n!

 

Note that  every term in  red  will have 3   as a factor because we can write

 

( 1! + 2!)  +  (1 * 2 * 3) +  (1 * 2 * 3 * 4) + ......+ (1 * 2 * 3 *...* (n - 2) )  + 

(1 * 2 * 3 * ... * (n - 1) )  +  ( 1 * 2 * 3 * ..... n )

 

So  the first term  = 3 and the sum of the rest of the terms can be written  as  3a   since  3 is a factor of each

 

So....the sum becomes   3  +  3a  =  3 (1  + a) which is divisible by 3

 

 

cool cool cool

CPhill  Feb 3, 2018
edited by CPhill  Feb 3, 2018
 #2
avatar+11847 
+1

Nice answer CPhill...i just dont get the par twhere you used "a"....what does that mena?could you write it in a longer form....

 

my teacher's answer is in this way:

 

1!+2!+3!+4!+5!/3

 

=3+3!+4!+5!/3

 

+3(1+ 1x2 + 1x2x4 + 1x2x4x5)/3

 

3!, 4! .......n! all have a factory of 3

 

1!+2!=3

 

therefore, ....... is divisible by 3.

 

_______________________________________________

 

i asked him about the answer and he said that those added terms will extend to n , tho i dont understand how to write them by taking 3 as a factor,.....could you pls solve it like that.!!!! Thank you so much CPhill....

 

 

Heres an energy frink for you....to help you answer my questions!thanks.

rosala  Feb 5, 2018

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