+0  
 
0
158
2
avatar+11847 

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
Sort: 

2+0 Answers

 #1
avatar+85757 
+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

30 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details