bn vào câu hỏi tương tự có nha!

chúc hok tốt

\(\frac{1}{4}=\frac{1}{2.2}< \frac{1}{1.2}=\frac{1}{2}-\frac{1}{4}\)

\(\Leftrightarrow\frac{1}{16}< \frac{1}{2.4}=\frac{1}{4}-\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{36}< \frac{1}{4.6}=\frac{1}{8}-\frac{1}{12}\)

\(\Leftrightarrow\frac{1}{64}< \frac{1}{6.8}=\frac{1}{12}-\frac{1}{16}\)

\(\Leftrightarrow\frac{1}{100}< \frac{1}{8.10}=\frac{1}{16}-\frac{1}{20}\)

\(\Leftrightarrow\frac{1}{144}< \frac{1}{10.12}=\frac{1}{20}-\frac{1}{24}\)

\(\Leftrightarrow\frac{1}{196}< \frac{1}{12.14}=\frac{1}{24}-\frac{1}{28}\)

\(\Rightarrow\frac{1}{4}+\frac{1}{16}+.....+\frac{1}{196}< \frac{1}{2}-\frac{1}{28}< \frac{1}{2}ĐPCM\)

\(4B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{7^2}\)

Ta lại có: \(4B-1\le\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{6.7}=1-\frac{1}{7}=\frac{6}{7}

minh cung chiu

ai làm đc

P = 1/4 + 1/16 + 1/36 + .. + 1/196 = 1/2² + 1/4² + 1/6² +...+ 1/12² + 1/14² 

xét tổng quát với số nguyên dương k ta có: 
(2k-1)(2k+1) = 4k² - 1 < 4k² = (2k)² => 1/(2k)² < 1/(2k-1)(2k+1) 
=> 2/(2k)² < 2 /(2k-1)(2k+1) = 1/(2k-1) - 1/(2k+1) (*) 

ad (*) cho k từ 1 đến 7 
2/2² < 1/1 - 1/3 
2/4² < 1/3 - 1/5 
... 
2/12² < 1/11 - 1/13 
2/14² < 1/13 - 1/15 
+ + cộng lại + + 
2/2² + 2/4² +...+ 2/14² < 1/1 - 1/15 < 1 
=> 2(1/2² + 1/4² +..+ 1/14²) < 1 => P < 1/2 (đpcm)

\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)

= \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

= \(\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)< \frac{1}{4}\left(1+1\right)=\frac{1}{2}\)

#ĐinhBa

B=1/2²+1/4²+...+1/14²

4B=1+1/2²+...+1/7² nhỏ hơn hoặc bằng 1+1/1.2+1/2.3+...+1/6.7 = 2-1/7=13/7

B nhỏ hỏn hoặc bằng 13/28 nhỏ hơn 1/2

vậy B nhỏ hơn 1/2

\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)

\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{7^2}\right)\)

\(< \frac{1}{2^2}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{1}{2^2}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2^2}\cdot\frac{6}{7}\)

\(=\frac{3}{14}\)

\(< \frac{1}{2}\)

\(\frac{1}{16}\)<\(\frac{1}{3\cdot4}\)tương tự=>\(\frac{1}{4}+\)\(\frac{1}{16}\)+.......+\(\frac{1}{196}< \frac{1}{3\cdot4}+......+\frac{1}{8\cdot9}=\frac{1}{3}\)--\(\frac{1}{9}\)+\(\frac{1}{4}\)=\(\frac{7}{18}< \frac{1}{2}\)

Vậy.................