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# prove that is divisible by 3. (n should be greater than or equal to 2)

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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)

Feb 3, 2018

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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

Feb 3, 2018
edited by CPhill  Feb 3, 2018
#2
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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.

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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....