Giải thích các bước giải:

Ta có:

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

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

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

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

$\to A=(\dfrac11+\dfrac13+...+\dfrac1{49})+(\dfrac12+\dfrac14+...+\dfrac1{50})-2(\dfrac12+\dfrac14+...+\dfrac1{50})$

$\to A=(\dfrac11+\dfrac12+\dfrac13+..+\dfrac1{49}+\dfrac1{50})-(\dfrac11+\dfrac12+..+\dfrac1{25})$

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

Đáp án:

Giải thích các bước giải:

Chứng minh rằng :

$\frac{1}{1.2}$ + $\frac{1}{3.4}$ + $\frac{1}{5.6}$ + ... + $\frac{1}{49.50}$ = $\frac{1}{26}$ + $\frac{1}{27}$ + $\frac{1}{28}$ + ...+$\frac{1}{50}$ 

= $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + $\frac{1}{5}$ - $\frac{1}{6}$ + ... + $\frac{1}{49}$ - $\frac{1}{50}$.

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

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

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

= $\frac{1}{26}$ + $\frac{1}{27}$ + $\frac{1}{28}$ + ... + $\frac{1}{50}$ 

Vậy $\frac{1}{1.2}$ + $\frac{1}{3.4}$ + $\frac{1}{5.6}$ + ... + $\frac{1}{49.50}$ = $\frac{1}{26}$ + $\frac{1}{27}$ + $\frac{1}{28}$ + ...+$\frac{1}{50}$