Bài 5 (0,5 điểm): Cho biểu thức A =$\frac{1}{1.2}$ + $\frac{1}{3.4}$ + $\frac{1}{5.6}$ +...+ $\frac{1}{49.50}$. Chứng minh rằng A<1

Giải thích các bước giải:

Ta có:

$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac1{49.50}$

$\to A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+\dfrac{6-5}{5.6}+...+\dfrac{50-49}{49.50}$

$\to A=\dfrac11-\dfrac12+\dfrac13-\dfrac14+\dfrac15-\dfrac16+...+\dfrac1{49}-\dfrac1{50}$

$\to A=\left(\dfrac11+\dfrac13+\dfrac15+...+\dfrac1{49}\right)-\left(\dfrac12+\dfrac14+...+\dfrac1{50}\right)$

$\to A=\left(1+\dfrac13+\dfrac15+...+\dfrac1{49}\right)-\left(\dfrac12+\dfrac14+...+\dfrac1{50}\right)$

$\to A=\left(1+\dfrac13+\dfrac15+...+\dfrac1{49}\right)+\left(\dfrac12+\dfrac14+...+\dfrac1{50}\right)-2\left(\dfrac12+\dfrac14+...+\dfrac1{50}\right)$

$\to A=\left(1+\dfrac12+\dfrac13+\dfrac14+\dfrac15+....+\dfrac1{49}+\dfrac1{50}\right)-\left(1+\dfrac12+...+\dfrac1{25}\right)$

$\to A=\dfrac1{26}+\dfrac1{27}+\dfrac{1}{28}+...+\dfrac{1}{50}$

$\to A<\dfrac1{25}+\dfrac1{25}+\dfrac{1}{25}+...+\dfrac{1}{25}$

$\to A<25\cdot\dfrac1{25}$

$\to A<1$