TA CÓ:
Đặt C = \(\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}\)
\(\Rightarrow\frac{2}{5}C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{29.31}\)
\(\Rightarrow\frac{2}{5}C=\frac{1}{3}-\frac{1}{31}=\frac{28}{93}\)
\(\Rightarrow C=\frac{28}{93}:\frac{2}{5}=\frac{70}{93}\)
Tương tự, đặt \(D=\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}\)
Mà \(D=C\)
\(\Rightarrow D=\frac{70}{93}\)
Đặt \(B=\frac{3}{2.3}+\frac{3}{3.4}+.....+\frac{3}{15.16}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{15.16}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{2}-\frac{1}{16}=\frac{7}{16}\)
\(\Rightarrow B=\frac{7}{16}:\frac{1}{3}=\frac{21}{16}\)
Nên \(\frac{5}{3.5}+...+\frac{5}{29.31}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{29.31}+\frac{3}{2.3}+..+\frac{3}{15.16}\)
\(\Rightarrow C+D+B=C.2+B=\frac{140}{93}+\frac{21}{16}=2,81\)
