Toán 7 trường chuyên - khó

G

girlhay99

Last edited by a moderator:
I

icy_tears

Ta có:
$\frac{1}{199} + \frac{2}{198} + \frac{3}{197} + ... + \frac{199}{1}$
$= \frac{1}{199} + \frac{2}{198} + \frac{3}{197} + ... + \frac{199}{1} + 199 - 199$
$= \frac{1}{199} + 1 + \frac{2}{198} + 1 + \frac{3}{197} + 1 + ... + \frac{199}{1} + 1 - 199$
$= \frac{200}{199} + \frac{200}{198} + \frac{200}{197} + ... + \frac{200}{1} - 199$
$= 200(\frac{1}{199} + \frac{1}{198} + \frac{1}{197} + ... + \frac{1}{2}) + 200 - 199$
$= 200(\frac{1}{199} + \frac{1}{198} + \frac{1}{197} + ... + \frac{1}{2}) + 1$
$= 200(\frac{1}{200} + \frac{1}{199} + \frac{1}{198} + \frac{1}{197} + ... + \frac{1}{2})$

\Rightarrow $\frac{\frac{1}{199} + \frac{2}{198} + \frac{3}{197} + ... + \frac{199}{1}}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{200}}$
$= \frac{200(\frac{1}{200} + \frac{1}{199} + \frac{1}{198} + \frac{1}{197} + ... + \frac{1}{2})}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{200}}$
$= 200$

 
Last edited by a moderator:
Top Bottom