Question

a , A= \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{99.101}\)_ \(\dfrac{1}{101}\)

Tính A

Answer 1

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

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

\(A=\dfrac{1}{2}.\left(\dfrac{100}{101}\right)-\dfrac{1}{101}\)

\(A=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)

Answer 2

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}-\dfrac{1}{101}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)-\dfrac{1}{101}\)

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

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

\(=\dfrac{1}{2}.\dfrac{100}{101}-\dfrac{1}{101}=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)

Vậy...

Answer 3

Giải:

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}-\dfrac{1}{101}\)

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

\(\Leftrightarrow A=\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{101}\right)-\dfrac{1}{101}\)

\(\Leftrightarrow A=\dfrac{1}{2}.\dfrac{100}{101}-\dfrac{1}{101}\)

\(\Leftrightarrow A=\dfrac{50}{101}-\dfrac{1}{101}=\dfrac{49}{101}\)

Vậy \(A=\dfrac{49}{101}\)

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