How to prove this inequality by geometric methods?
$$2500 \pi-100<\sqrt{1 \cdot 199}+\sqrt{2 \cdot 198}+\cdots+\sqrt{99 \cdot 101}<2500\pi$$
How to prove this inequality by geometric methods?
$$2500 \pi-100<\sqrt{1 \cdot 199}+\sqrt{2 \cdot 198}+\cdots+\sqrt{99 \cdot 101}<2500\pi$$
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Note that $$\sqrt{k(200-k)}=\sqrt{100^2-(k-100)^2}\qquad(1\leq k\leq99)\ .$$ This means that you should look at the circular disc $(x-100)^2+y^2\leq100^2$. Your sum is a rectangle approximation to the area of the upper left quarter of this disc.