Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+..+\dfrac{1}{49.50}$
$\to A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+..+\dfrac{50-49}{49.50}$
$\to A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{49}-\dfrac{1}{50}$
$\to A=(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{49})-(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{50})$
$\to A=(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{50})-2(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{50})$
$\to A=(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50})-(\dfrac{1}{1}+\dfrac{1}{2}+..+\dfrac{1}{25})$
$\to A=\dfrac{1}{26}+\dfrac{1}{27}+..+\dfrac{1}{50}$
$\to 4A=\dfrac{4}{26}+\dfrac{4}{27}+..+\dfrac{4}{50}$
$\to 4A=(\dfrac{4}{26}+1)+(\dfrac{4}{27}+1)+..+(\dfrac{4}{50}+1)-25$
$\to 4A=\dfrac{30}{26}+\dfrac{31}{27}+..+\dfrac{54}{50}-25$
$\to C=A:(4A)=\dfrac{1}{4}$
