Question

Tính:

A=\(\dfrac{1}{3\cdot8}+\dfrac{1}{6\cdot12}+\dfrac{1}{9\cdot16}+...+\dfrac{1}{1512\cdot2020}\)

Answer 1

\(A=\dfrac{1}{3.4}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{504.505}\right)=\)

\(=\dfrac{1}{3.4}\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{505-504}{504.505}\right)=\)

\(=\dfrac{1}{3.4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{504}-\dfrac{1}{505}\right)=\)

\(=\dfrac{1}{3.4}\left(1-\dfrac{1}{505}\right)=\dfrac{1}{3.4}.\dfrac{504}{505}=\dfrac{42}{505}\)