Question

[Nguyễn Võ Ngọc Ngân]

\(A=\frac{3}{1^2x2^2}+\frac{5}{2^2x3^2}+\frac{7}{3^2x4^2}+......+\frac{39}{19^2x20^2}\)

Tính A

Answer 1

[Lê Tài Bảo Châu]

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{39}{19^2.20^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{19^2}-\frac{1}{20^2}\)

\(=1-\frac{1}{20^2}\)

\(=\frac{399}{400}\)