Ta nhận thấy
\(\dfrac{1}{n\cdot\left(n+2\right)}-\dfrac{1}{\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{n+4}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}-\dfrac{n}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{n+4-n}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\\ =\dfrac{4}{n\cdot\left(n+2\right)\cdot\left(n+4\right)}\)
\(A=\dfrac{4}{2\cdot4\cdot6}+\dfrac{4}{4\cdot6\cdot8}+\dfrac{4}{6\cdot8\cdot10}+...+\dfrac{4}{46\cdot48\cdot50}\\ =\dfrac{1}{2\cdot4}-\dfrac{1}{4\cdot6}+\dfrac{1}{4\cdot6}-\dfrac{1}{6\cdot8}+\dfrac{1}{6\cdot8}-\dfrac{1}{8\cdot10}+...+\dfrac{1}{46\cdot48}-\dfrac{1}{48\cdot50}\\ =\dfrac{1}{2\cdot4}-\dfrac{1}{48\cdot50}\\ =\dfrac{1}{8}-\dfrac{1}{2400}\\ =\dfrac{300}{2400}-\dfrac{1}{2400}\\ =\dfrac{299}{2400}\)
Số nghịch đảo của \(A\) là \(\dfrac{2400}{299}\)
