Miquel-point wrote: For each $n\in \mathbb{Z}_{>0}$ let $a_n$ be the closest integer to $\sqrt{n}$. Compute the general term of the sequence: $(b_n)_{n\geqslant 1}$ with \[b_n=\sum_{k=1}^{n^2} a_k.\] Note that $\forall k\in[n^2-n+1,n^2+n]$ : $a_k=n$ So $\sum_{k=1}^{n^2}a_k=\sum_{m=1}^{n-1}\sum_{k=m^2-m+1}^{m^2+m}a_k+\sum_{k=n^2-n+1}^{n^2}a_k$ $=\sum_{m=1}^{n-1}\sum_{k=m^2-m+1}^{m^2+m}m+\sum_{k=n^2-n+1}^{n^2}n$ $=\sum_{m=1}^{n-1}2m^2+n^2$ $=\frac{(n-1)n(2n-1)}3+n^2$ $=\boxed{\frac{2n^3+n}3}$