Tìm n biết $\frac{1}{8}$+$\frac{1}{24}$+$\frac{1}{48}$+…+$\frac{1}{9800}$= $\frac{(n+5^{2})}{200}$ 22/11/2021 Bởi Brielle Tìm n biết $\frac{1}{8}$+$\frac{1}{24}$+$\frac{1}{48}$+…+$\frac{1}{9800}$= $\frac{(n+5^{2})}{200}$
$\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+…+\frac{1}{9800}=\frac{n+5^2}{200}$ $⇔\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+…+\frac{1}{98.100}=\frac{n+25}{200}$ $⇔\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+…+\frac{2}{98.100}=\frac{2.(n+25)}{200}$ $⇔\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+…+\frac{1}{98}-\frac{1}{100}=\frac{n+25}{100}$ $⇔\frac{1}{2}-\frac{1}{100}=\frac{n+25}{100}$ $⇔\frac{49}{100}=\frac{n+25}{100}$ $⇔49=n+25$ $⇔n=24$ Vậy $n=24$. Bình luận
$\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+…+\frac{1}{9800}=\frac{n+5^2}{200}$
$⇔\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+…+\frac{1}{98.100}=\frac{n+25}{200}$
$⇔\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+…+\frac{2}{98.100}=\frac{2.(n+25)}{200}$
$⇔\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+…+\frac{1}{98}-\frac{1}{100}=\frac{n+25}{100}$
$⇔\frac{1}{2}-\frac{1}{100}=\frac{n+25}{100}$
$⇔\frac{49}{100}=\frac{n+25}{100}$
$⇔49=n+25$
$⇔n=24$
Vậy $n=24$.