b)
Ta có :
$A= \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$
$= (\frac{1}{1.2}+\frac{1}{3.4}) + \frac{1}{5.6}+...+\frac{1}{99.100}$
$= \frac{7}{12} + \frac{1}{3.4}) + \frac{1}{5.6}+...+\frac{1}{99.100} > \frac{7}{12} (1)$
Lại có :
$A = \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$
$= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{99} - \frac{1}{100}$
$= (1 - \frac{1}{2} + \frac{1}{3}) - \frac{1}{4} + ... + \frac{1}{99} - \frac{1}{100}$
$= \frac{5}{6} - \frac{1}{4} + ... + \frac{1}{99} - \frac{1}{100}$
$= \frac{5}{6} - (\frac{1}{4} - ... - \frac{1}{99} + \frac{1}{100}) < \frac{5}{6} (2)$
Từ $(1)$ và $(2)$
$\Rightarrow \frac{7}{12} < A < \frac{5}{6}$
a) $A= \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$
$= (\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6} + ... + \frac{1}{49.50}) + (\frac{1}{51.52} + \frac{1}{53.54} + \frac{1}{55.56} + ... + \frac{1}{99.100})$
$= (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{49} - \frac{1}{50}) + (\frac{1}{51} - \frac{1}{52} + \frac{1}{53} - \frac{1}{54} + ... + \frac{1}{99} - \frac{1}{100})$
$= [(1 + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{49}) - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{50})] + [(\frac{1}{51} + \frac{1}{53} + \frac{1}{55} + ... \frac{1}{99}) - (\frac{1}{52} + \frac{1}{54} + \frac{1}{56} + ... + \frac{1}{100})]$
$= [(1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{49} + \frac{1}{50}) - 2(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{50})] + [(\frac{1}{51} + \frac{1}{52} + \frac{1}{53} + ... \frac{1}{99} + \frac{1}{100}) - 2(\frac{1}{52} + \frac{1}{54} + \frac{1}{56} + ... + \frac{1}{100})]$
$= (\frac{1}{26} + \frac{1}{27} + \frac{1}{28} + ... + \frac{1}{50}) + \frac{1}{51} + \frac{1}{52} + \frac{1}{53} + ... \frac{1}{100}) - \frac{1}{26} + \frac{1}{27} + \frac{1}{28} + ... + \frac{1}{50})$
$= \frac{1}{51} + \frac{1}{52} + \frac{1}{53} + ... \frac{1}{100}$
Em thấy anh giải thế này chưa đúng lắm và đang còn dài dòng nên e giải lại:
a) Ta có : $A= \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}$
$\Leftrightarrow$$A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}$
$\Leftrightarrow$$A=(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100})$
$\Leftrightarrow$$A=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100})-2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100})$
$\Leftrightarrow$$A=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{100})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{50})$
$\Rightarrow A= \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$
b) Ta có: $A= \frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}$
$\Leftrightarrow$$A=(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75})+(\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100})$
$\Leftrightarrow$$A>\frac{1}{75}.25+\frac{1}{100}.25$
$\Leftrightarrow$$A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}(1)$
Mặt khác:
$A<\frac{1}{50}.25+\frac{1}{75}.25$
$\Leftrightarrow$$A<\frac{1}{2}+\frac{1}{3}=\frac{5}{6}(2)$
Từ (1) và (2) $\Rightarrow \frac{7}{12}<A<\frac{5}{6}$
Bài viết đã được chỉnh sửa nội dung bởi ngocsangnam15: 29-08-2015 - 20:33