`=>` Tặng bạn
`a)`
$\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)$
$\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)$
$\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)$
$\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)$
$\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\) $
`b)` Bạn cho số cuối là 2n
Rồi cm tt nhá