Question

Chứng minh

\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{49\cdot50}=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\)

Answer 1

Lời giải:

\(\text{VT}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{49.50}\)

\(=\frac{2-1}{1.2}+\frac{4-3}{3.4}+....+\frac{50-49}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}....+\frac{1}{50}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}-\left(1+\frac{1}{2}+\frac{1}{3}....+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{49}+\frac{1}{50}\)

Do đó ta có đpcm.