Tính tổng: S = 1-1/2 + 1/3-1/4 + 1/5-1/6 + ..... 1/499-1/500

* Hướng dẫn giải

Tính tổng:

S = 1$-$$\frac{1}{2}$ + $\frac{1}{3}$$-$$\frac{1}{4}$ + $\frac{1}{5}-\frac{1}{6}$+ ..... $\frac{1}{499}-\frac{1}{500}$S = 1$-\frac{1}{2}$ + $\frac{1}{3}-\frac{1}{4}$ + $\frac{1}{5}-\frac{1}{6}$ + ..... $\frac{1}{499}-\frac{1}{500}$ = (1 + $\frac{1}{3}$ + $\frac{1}{5}$ + ..+ $\frac{1}{499}$) $-$ ($\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$ + ...+ $\frac{1}{500}$) $-$($\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$ + ...+ $\frac{1}{500}$) + ($\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$ + ...+ $\frac{1}{500}$) S = (1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ....+ $\frac{1}{500}$) $-$ 2.($\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$ + ...+ $\frac{1}{500}$) = (1 + $\frac{1}{2}$ + $\frac{1}{3}$ + $\frac{1}{4}$ + ....+ $\frac{1}{500}$)$-$ (1 + $\frac{1}{2}$ + $\frac{1}{3}$ + ...+$\frac{1}{250}$) = $\frac{1}{251}$ + $\frac{1}{252}$ + ...+ $\frac{1}{500}$.

Vậy S = $\frac{1}{251}$ + $\frac{1}{252}$+ ...+ $\frac{1}{500}$