Toán Giúp mình với! 1/1*2*3*4+1/2*3*4*5+1/3*4*5*6+…+1/47*48*49*50 12/09/2021 By Allison Giúp mình với! 1/1*2*3*4+1/2*3*4*5+1/3*4*5*6+…+1/47*48*49*50
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 \(A=\dfrac{6533}{117600}\) Xin hay nhất Trả lời
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 \(A=\dfrac{6533}{117600}\)
Xin hay nhất