Question

Tính A , biết :

\(A=\dfrac{2}{11.15}+\dfrac{2}{15.19}+\dfrac{2}{19.23}+...+\dfrac{2}{51.55}\)

Answer 1

Lời giải:

\(A=\frac{2}{11.15}+\frac{2}{15.19}+\frac{2}{19.23}+...+\frac{2}{51.55}\)

\(\Rightarrow 2A=\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+...+\frac{4}{51.55}\)

\(=\frac{15-11}{11.15}+\frac{19-15}{15.19}+\frac{23-19}{19.23}+....+\frac{55-51}{51.55}\)

\(=\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+...+\frac{1}{51}-\frac{1}{55}\)

\(=\frac{1}{11}-\frac{1}{55}=\frac{4}{55}\)

\(\Rightarrow A=\frac{2}{55}\)