A = \(\dfrac{7}{2.3}+\dfrac{7}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{7}{48.49.50}\)
A = \(7.\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{48.49.50}\right)\)
\(\dfrac{A}{7}\) = \(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{48.49.50}\right)\)
Bạn để ý kĩ phần này:
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\)
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{2}{2.3.4}\)
\(\dfrac{1}{3.4}-\dfrac{1}{4.5}=\dfrac{2}{3.4.5}\)
...
\(\dfrac{1}{48.49}-\dfrac{1}{49.50}=\dfrac{2}{48.49.50}\)
Qua đó, bạn có thể rút ra công thức tổng quát sau:
\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
Từ đó suy ra:
\(\dfrac{2A}{7}\) = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{48.49.50}\)
\(\dfrac{2A}{7}=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50}\)
\(\dfrac{2A}{7}=\dfrac{1}{2}-\dfrac{1}{49.50}\)
\(\dfrac{2A}{7}=\dfrac{612}{1225}\)
\(2450A=4284\)
A = \(\dfrac{306}{175}\)
mình làm sai rồi bạn, làm lại này:
\(\dfrac{7}{1.2.3}+\dfrac{7}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{7}{48.49.50}\)
\(=\dfrac{7}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{48.49.50}\right)\)
\(=\dfrac{7}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50}\right)\)
\(=\dfrac{7}{2}\left(\dfrac{1}{2}-\dfrac{1}{49.50}\right)=\dfrac{7}{4}-\dfrac{1}{7.100}=\dfrac{306}{175}\)
\(\dfrac{7}{1.2.3}+\dfrac{7}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{7}{48.49.50}=7\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{48.49.50}\right)\)
\(=7\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50}\right)\)
\(=7\left(\dfrac{1}{1.2}-\dfrac{1}{49.50}\right)=\dfrac{7}{2}-\dfrac{1}{7.50}=\dfrac{1225}{350}-\dfrac{1}{350}=\dfrac{1224}{350}=\dfrac{612}{175}\)
