Question

[Thái Hoàng Thục Anh]

A = (1/2^2 -1).(1/3^2 -1)(1/4^2 -1)....(1/100^2 -1)

so sáh A vs -1/2

Answer 1

[robert lewandoski]

ta có:

\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+..+\frac{19}{9^2.10^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+..+\frac{10^2-9^2}{9^2.10^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+..+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}