\(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}\)
\(=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)
\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Xét \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)>49-1=48\)
Vậy \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\)(đpcm)
