Đặt A=\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{49.51}\)
\(\Rightarrow2A=2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{49.51}\right)\)
\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{49.51}\)
\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{49}-\dfrac{1}{51}\)
(do \(\dfrac{n}{a.\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với mọi \(a\in N\)*)
\(\Rightarrow2A=1-\dfrac{1}{51}\Rightarrow2A=\dfrac{50}{51}\Rightarrow A=\dfrac{25}{51}\)
Vậy \(A=\dfrac{25}{51}\)
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Gọi A = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(\Rightarrow2A=\)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)
\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{51}=\dfrac{50}{51}\)
\(\Rightarrow A=\dfrac{50}{51}:2=\dfrac{25}{51}\)
Vậy A = 0,(4901960784313725)
Sai đề, đề đúng và cách làm như sau:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}\right)+\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}.\dfrac{50}{51}=\dfrac{25}{51}.\)
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}.............+\dfrac{1}{49.51}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+........+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{51}\right)\)
\(=\dfrac{1}{2}.\dfrac{50}{51}=\dfrac{50}{102}\)
