Câu trả lời:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)

 

\(\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-....-\frac{1}{50}+\frac{1}{50}\)

=> \(\frac{1}{2}-\frac{1}{50}=\frac{12}{25}\)