# novice question

• Jun 25th 2008, 07:26 PM
juneman
novice question
Why do the sum of all numbers in any square / sum of square diagonal numbers = the square root?
123
456
789
1+2+3+4+5+6+7+8+9 = 45
1+5+9 or 3+5+7 = 15
• Jun 26th 2008, 12:04 AM
red_dog
$\begin{array}{cccccc}1 & 2 & 3 & \ldots & n-1 & n\\
n+1 & n+2 & n+3 & \ldots & 2n-1 & 2n\\
2n+1 & 2n+2 & 2n+3 & \ldots & 3n-1 & 3n\\
\dots & \ldots & \ldots & \ldots & \ldots & \ldots \\
(n-2)n+1 & (n-2)+2 & (n-2)n+3 & \ldots & (n-2)n+n-1 & (n-1)n\\
(n-1)n+1 & (n-1)n+2 & (n-1)n+3 & \ldots & n^2-1 & n^2
\end{array}$

The sum of all numbers is $S_1=1+2+3+\ldots+n^2=\frac{(1+n^2)n^2}{2}$

The sum of diagonal numbers is
$S_2=1+(n+2)+(2n+3)+\ldots+[(n-2)n+n-1]+n^2=$
$=(1+2+3+\ldots+(n-1))+n(1+2+3+\ldots+(n-2))+n^2=\frac{(1+n^2)n}{2}$

Now $\frac{S_1}{S_2}=n=\sqrt{n^2}$