Question

A=\(\dfrac{6}{3}\)+\(\dfrac{6}{15}\)+\(\dfrac{6}{35}\)+...+\(\dfrac{6}{9999}\)

Answer 1

\(A=\dfrac{6}{3}+\dfrac{6}{15}+\dfrac{6}{35}+...+\dfrac{6}{9999}\)

\(=3\left(\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\right)\)

\(=3\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=3\left(1-\dfrac{1}{101}\right)=3\cdot\dfrac{100}{101}=\dfrac{300}{101}\)

Answer 2

ai biết làm giúp mình với