How do you prove that the sum of 1 to k divided by k is a multiple of 1/2? Expressed mathematically: $\frac{1}{k}\sum_{i=1}^k n = \frac{x}{2}$ where x is an integer. I tested this series mathematically and I found this trend, but I don't know how to actually prove it. Thanks in advance. 
 A: It is an easy and well-known result that 
$$
\sum_{i=1}^k i=\frac{k(k+1)}2. 
$$
For a not very formal but very intuitive proof, write 
$$
\begin{array}{rccccccccc}
S&=&1&+&2&+&3&+&\cdots&+&k-1&+&k\\
S&=&k&+&(k-1)&+&\cdots&+&3&+&2&+&1
\end{array}
$$
If you now add both rows, you get 
$$
2S=(k+1)+(k+1)+\cdots+(k+1)=k(k+1).
$$
A: Notice that $\frac 1k \sum\limits_{n=1}^k n$ is, by definition the average of $1,2,3,4......, .... k$.  Convince yourself that the average of $1....n$ (equally distributed) is $\frac {k+1}2$.
.... or ...
Let $N =  \sum\limits_{n=1}^k n = \sum\limits_{n=k;-1}^1 n = \sum\limits_{n=1}^k ((k+1) -n)$
So $2N =  (\sum\limits_{n=1}^k n) + (\sum\limits_{n=1}^k ((k+1) -n))$
$= \sum\limits_{n=1}^k [n + (k=1) - n] = \sum\limits_{n=1}^k (k+1)$
$= k*(k+1)$
So $N = \frac {k(k+1)}{2}$
.... or ....
$N = 1+2+3..........+ (k-2) + (k-1) + k$
$N = k + (k-1) + (k-2) + .....+ 3 + 2 + 1$
$N+N = (1+k) + (2+k-1) + (3k-2 ) + ........ +(k-2+3) + (k-1+2) + (k+1)$
$2N = (k+1) + (k+1) + (k+1) + ..... + (k+1) + (k+1) + (k+1)$
$2N = k*(k+1)$.
So $N = \frac {k(k+1)}2$
..... or do induction.....
$\frac 1k \sum^k n = \frac {k+1}2$ is true for $k = 1$ as $\frac 11 = \frac 22$.
$\frac 1k \sum^k n = \frac {k+1}2$ then
$\sum^k n = \frac {k(k+1)}2$
$\sum^{k+1}n = \frac {k(k+1)}2 + (k+1)$
$= \frac {k(k+1)}2 +\frac {2(k+1)}2$
$= \frac {(k+2)(k+1)}2$
So $\frac {1}{k+1} \sum^{k+1}n = \frac {k+2}2$.
