Question

\(\frac{1}{255}+\frac{1}{323}+\frac{1}{399}+......+\frac{1}{9999}\)

Answer 1

\(\frac{1}{255}+\frac{1}{323}+...+\frac{1}{9999}\)

=\(\frac{1}{15.17}+\frac{1}{17.19}...+\frac{1}{99.101}\)

=\(\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

=\(\frac{1}{15}-\frac{1}{101}\)

= \(\frac{86}{1515}\)

Xong roài đó bạn

Answer 2

Đặt \(A=\frac{1}{225}+\frac{1}{323}+\frac{1}{399}+....+\frac{1}{9999}\)

\(A=\frac{1}{15.17}+\frac{1}{17.19}+\frac{1}{19.21}+...+\frac{1}{99.101}\)

\(2A=\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}+...+\frac{1}{99}-\frac{1}{101}\)

\(2A=\frac{1}{15}-\frac{1}{101}=\frac{86}{1515}\)

\(\Rightarrow A=\frac{86}{1515}\div2=\frac{43}{1515}\)