共通テストとかで出てきそうなやつを言い換えただけ
Sを求めよ.
$$S=\sum_{k=1}^n\floor{\sqrt{k}}$$
$$\begin{align} S&=\sum_{k=1}^n\sum_{m=1}^\floor{\sqrt{k}}1\\ &=\sum_{1\leq m^2\leq k \leq n}1\\ &=\sum_{m=1}^\floor{\sqrt{n}}(n-m^2+1)\\ &=\floor{\sqrt{n}}(n+1)-\frac{\floor{\sqrt{n}}(\floor{\sqrt{n}}+1)(2 \floor{\sqrt{n}}+1)}6\\ \therefore S&=\frac{\floor{\sqrt{n}}}6(6n-2\floor{\sqrt{n}}^2-3 \floor{\sqrt{n}}+5)\\ \end{align}$$
なにこれ