Chứng minh rằng: 1/1.2 + 1/3.4 + 1/5.6 + ...+1/49.50

* Hướng dẫn giải

Ta có:

$\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\mathrm{...}+\frac{1}{49.50}$

$\begin{array}{l}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots \cdots +\frac{1}{49}-\frac{1}{50}\\ =\left(1+\frac{1}{3}+\frac{1}{5}+\cdots \cdots +\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots \cdots +\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots \cdots +\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots \cdots +\frac{1}{50}\right)\\ =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots \cdots +\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots \cdots +\frac{1}{25}\right)\\ =\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\mathrm{...}+\frac{1}{50}\left(dpcm\right)\end{array}$