Question

[Nguyễn Linh Ngọc]

A=\(\frac{1}{2.3}\)+\(\frac{1}{6.5}\)+\(\frac{1}{10.7}\)+...+\(\frac{1}{198.101}\)

Mik đang cần gấp.Ai giải nhanh mik tick cho nha

Answer 1

[Thanh Hà]

\(A=\frac{1}{2.1.3}+\frac{1}{2.3.5}+\frac{1}{2.5.7}+...+\frac{1}{2.99.101}\)

\(=\frac{1}{2}.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

Answer 2

[QuocDat]

\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+...+\frac{1}{198.101}\)

Đặt :\(M=\frac{1}{2.6}+\frac{1}{6.10}+...+\frac{1}{194.198}\)

\(M=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{194}-\frac{1}{198}\right)\)

\(M=\frac{1}{4}\left(\frac{1}{2}-\frac{1}{198}\right)\)

\(M=\frac{1}{4}.\frac{49}{99}\)

\(M=\frac{49}{396}\)

Đặt \(N=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)

\(N=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(N=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(N=\frac{1}{2}.\frac{98}{303}\)

\(N=\frac{49}{303}\)

Vậy ta có : A = M + N = \(\frac{49}{396}+\frac{49}{303}\) , bạn tự tính luôn nha