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**+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

CPhill
Feb 3, 2018

#2**+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.

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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....

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

rosala
Feb 5, 2018