Đáp án:
Giải thích các bước giải:
= ($\frac{1}{1.2}$+ $\frac{1}{3.4}$)+($\frac{1}{5.6}$+...+ $\frac{1}{99.100}$)
= $\frac{7}{12}$+ ($\frac{1}{5.6}$+...+ $\frac{1}{99.100}$)> $\frac{7}{12}$
Do $\frac{1}{5.6}$+...+ $\frac{1}{99.100}$> 0) (1)
+) B= ($\frac{1}{1.2}$+ $\frac{1}{3.4}$)+ ($\frac{1}{5.6}$+...+ $\frac{1}{99.100}$)
= 1- $\frac{1}{2}$+ $\frac{1}{3}$- $\frac{1}{4}$+ $\frac{1}{5}$- $\frac{1}{6}$+...+ $\frac{1}{99}$- $\frac{1}{100}$
= (1- $\frac{1}{2}$+ $\frac{1}{3}$)- ($\frac{1}{4}$- $\frac{1}{5}$)- ($\frac{1}{6}$- $\frac{1}{7}$- ....- ( $\frac{1}{98}$- $\frac{1}{99}$)- $\frac{1}{100}$
Do -($\frac{1}{4}$- $\frac{1}{5}$)- ($\frac{1}{6}$- $\frac{1}{7}$)- - ($\frac{1}{98}$- $\frac{1}{99}$)- $\frac{1}{100}$< 1- $\frac{1}{2}$+ $\frac{1}{3}$= $\frac{5}{6}$ (2)
Từ (1) và (2)⇒ $\frac{7}{12}$< A< $\frac{5}{6}$ $(đpcm)$
