The sum of integers in each row / column / major diagonal of a Magic Square of order $N$ is $\dfrac{N(N^2+1)}{2}$

Question

Prove that the sum of integers in each row / column / major diagonal of a Magic Square of order $N$ is equal to $$\dfrac{N(N^2+1)}{2}$$

I have tested the formula on Durer's "1514" magic square and it holds (where N=4).

Answer 1

Add all the numbers in the square. It's the sum from 1 to $n^2$, which is $n^2(n^2+1)/2$. This is equal to the sum of $n$ rows, so each row is $n(n^2+1)/2$.