$\sum\limits_{i=1}^{n^2}\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j = \sum\limits_{i=1}^{n^2}i+(i+1)+(i+1)+...+(i+n)$
Are these two sums equal? If n = 2 so the result of $\sum\limits_{i=1}^{n^2}\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j$ is equal 30 and for $\sum\limits_{i=1}^{n^2}i+(i+1)+(i+1)+...+(i+n)$ is equal 42 !
But in the book wrote two formula is equal! Please, description why if I have wrong!
I know the simpled formula is as: $$\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j=\sum\limits_{j=i}^{i+n}j= i+(i+1)+(i+1)+…+(i+n)$$
For the result 30 see : 
We have that
$$\sum\limits_{k=1}^j1=j$$
and then
$$\sum\limits_{j=i}^{i+n}\sum\limits_{k=1}^j=\sum\limits_{j=i}^{i+n}j= i+(i+1)+(i+1)+…+(i+n)$$
therefore the equality holds, you have probably made some mistake in the calculation.
We have that for $n=2$
$$\sum\limits_{i=1}^{4}\sum\limits_{j=i}^{i+2}\sum\limits_{k=1}^j = \sum\limits_{i=1}^{4}\sum\limits_{j=i}^{i+2}j= \sum\limits_{i=1}^{4} (i+i+1+i+2)=\sum\limits_{i=1}^{4}( 3i+3)=$$$$=3\sum\limits_{i=1}^{4}i+3\sum\limits_{i=1}^{4}1=30+12=42$$