Question 1175017
<pre>
One way to get a formula is to add 1+2+...+n to itself, but write the 2nd line in reverse, then add them:


Sn  =    1 +    2   +   3  + ... + (n-1) + n 
Sn  =  ( n + (n-1) + (n-2) + ... +    2  + 1 )
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2Sn = (n+1) + (n+1) + (n+1) + ... + (n+1) + (n+1)

The RHS is just n terms of n+1, thus it is n(n+1)

2Sn = n(n+1)

 {{{ S[n] = (1/2)n(n+1) }}}  


(i)
As a hint for divisibility by 5: 
      If n is even, n+1 is odd
      If n is odd, n+1 is even

Thus, a factor of 2 is always present in n(n+1), and the (1/2) in the {{{S[n]}}} formula will divide that out ("cancels" it).

So there is just the consideration about n or n+1 being divisible by 5.  Numbers ending with 5 or 0 meet this condition, and as long as EITHER n or n+1 ends with 5 or 0 you will find {{{S[n]}}} to be divisible by 5:  

Examples: 
n=10  >>>  n*(n+1)/2 = 110/2 = 55,  that works.     
n=9   >>>  n*(n+1)/2 = 9*10/2 = 45, that is divisible by 5
n=8   >>>  n*(n+1)/2 = 8*9/2 = 36, NOT divisible by 5 (neither n nor n+1 is divisble by 5)
n=15  >>>  n*(n+1)/2 = 15*16/2 = 120, divisible by 5
n=14  >>>  n*(n+1)/2 = 14*15/2 = 105, divisible by 5

A subtle note about the above: notice how the factor of 5 comes from 'n' or 'n+1' in a mutually exclusive way.  That is, if n has a factor of 5, n+1 does not;  if n+1 has a factor of 5, n does not.  You may need this observation for part(ii) when considering the factor 4.

(ii)
{{{S[n]}}} is even:  you give it some thought.  Remember, you always get one factor of 2 automatically in the product n(n+1) and that will cancel with the 1/2 of the formula, so the question is, for what values of n(n+1) do you get n(n+1) divisible by <b>4</b>?