Đáp án:
$\frac{6533}{117600}$
Giải thích các bước giải:
Đặt \(A=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{47.48.49.50}\)
Ta có:
\(3A=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{3}{3.4.5.6}+...+\dfrac{3}{47.48.49.50}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{47.48.49}-\dfrac{1}{48.49.50}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{48.49.50}\)
\(\Rightarrow3A=\dfrac{1}{6}-\dfrac{1}{117600}\)
\(\Rightarrow3A=\dfrac{6533}{39200}\)
\(\Rightarrow A=\dfrac{6533}{\dfrac{39200}{3}}=\dfrac{6533}{117600}\)
Vậy \(A=\dfrac{6533}{117600}\)
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