Question

[phamphuckhoinguyen]

Chứng minh rằng: (1/26+1/27+...+1/50)÷(1/1.2+1/3.4+...+1/49.50)=1

Answer 1

[Trí Tiên]

Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

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

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

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

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

Khi đó : \(\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)

\(=\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right):\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\right)=1\) (đpcm)

Answer 2

[Xyz OLM]

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

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

= \(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

Khi đó \(\frac{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}}=\frac{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}{\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}}=1\left(\text{đpcm}\right)\)