Lời giải:
Đặt \(A=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+.....+\frac{1}{\sqrt{2500}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{2500}}\)
\(\frac{A}{2}< \frac{1}{2}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{2499}+\sqrt{2500}}\)
\(\frac{A}{2}< \frac{1}{2}+\frac{\sqrt{2}-1}{(\sqrt{1}+\sqrt{2})(\sqrt{2}-\sqrt{1})}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+....+\frac{\sqrt{2500}-\sqrt{2499}}{(\sqrt{2499}+\sqrt{2500})(\sqrt{2500}-\sqrt{2499})}\)
\(\frac{A}{2}< \frac{1}{2}+(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{2500}-\sqrt{2499})\)
\(\frac{A}{2}< \frac{1}{2}+\sqrt{2500}-\sqrt{1}=49+\frac{1}{2}< 50\)
\(\Rightarrow A< 100\) (đpcm)
P.s: Bạn lưu ý lần sau gõ đề bài bằng công thức toán.
