khyeon wrote: $n \in \mathbb {N^+}$ Prove that the following equation can be expressed as a polynomial about $n$. $$\left[2\sqrt {1}\right]+\left[2\sqrt {2}\right]+\left[2\sqrt {3}\right]+ . . . +\left[2\sqrt {n^2}\right]$$ Let $f(n)=\sum_{k=1}^{n^2}\left\lfloor 2\sqrt k\right\rfloor$ $f(n+1)-f(n)=\sum_{k=n^2+1}^{n^2+2n+1}\left\lfloor 2\sqrt k\right\rfloor$ $f(n+1)-f(n)=\sum_{k=n^2+1}^{n^2+n}\left\lfloor 2\sqrt k\right\rfloor$ $+\sum_{k=n^2+n+1}^{n^2+2n}\left\lfloor 2\sqrt k\right\rfloor$ $+\left\lfloor 2\sqrt {(n+1)^2}\right\rfloor$ $f(n+1)-f(n)=n(2n)+n(2n+1)+(2n+2)$ $f(n+1)-f(n)=4n^2+3n+2$ Hence the claim.