Đặt: \(\left\{{}\begin{matrix}A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\\B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\end{matrix}\right.\)
Ta có:
\(B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{59}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{30}\right)\)
\(=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\)
\(\Rightarrow B=A\)
Vậy \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\) (Đpcm)
Ta có:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+......+\dfrac{1}{59.60}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{59}-\dfrac{1}{60}\)
= \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{59}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)
- \(2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)
= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{60}\right)\) - \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)+ \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)
- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)
= \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)
Vậy\(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{59.60}\)