Question

Tính giá trị biểu thức:

\(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)

Answer 1

Đặt \(A=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)

Ta có:

\(3A=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)

\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)

\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)

\(\Rightarrow3A=\dfrac{1}{6}-\dfrac{1}{24360}\)

\(\Rightarrow3A=\dfrac{1353}{8120}\)

\(\Rightarrow A=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)

Vậy \(A=\dfrac{451}{8120}\)

Answer 2

Ta có: \(\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)