Why $ \ \lfloor\sqrt n\rfloor= j(2j+1)$

Question

Can someone please explain to me why is this happening?

In general $\lfloor\sqrt n\rfloor$ is $j$ when $j^2\le n<(j+1)^2=j^2+2j+1$, which is $2j+1$ times, so if we group like terms together we find $$ S_k = \sum_{n=1}^{k^2-1} \lfloor\sqrt n\rfloor = \sum_{j=1}^{k-1} j(2j+1)$$

The only thing I don't understand is why $$ \ \lfloor\sqrt n\rfloor= j(2j+1)$$

Answer 1

This might be clearer: $$\sum_{n=1}^{k^2-1} \lfloor \sqrt n\rfloor = \sum_{j=1}^{k-1} \sum_{n=j^2}^{(j+1)^2-1} \lfloor \sqrt n\rfloor = \sum_{j=1}^{k-1} \sum_{n=j^2}^{(j+1)^2-1} j = \sum_{j=1}^{k-1} ((j+1)^2-j^2) j = \sum_{j=1}^{k-1} j (2j+1) $$