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ta có biêu thức trên\(\: < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)=\(\frac{2012}{2013}< 1\)
do dó biểu thức <1
Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{100^2}\)
Ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};\frac{1}{5^2}< \frac{1}{4.5};......;\frac{1}{100^2}< \frac{1}{99.100}\)
\(=>A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)
\(=>A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=>A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+.....+\frac{1}{100^2}< \frac{1}{2}\left(đpcm\right)\)
Bạn xem lời giải của mình nhé:
Giải:
Gọi \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\frac{1}{3^2}< \frac{1}{3.4}\\ \frac{1}{4^2}< \frac{1}{4.5}\\ ...\\ \frac{1}{100^2}< \frac{1}{99.100}\\ \Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{3}-\frac{1}{100}\\ \frac{1}{3}< \frac{1}{2}\Rightarrow\frac{1}{3}-\frac{1}{100}< \frac{1}{2}\\ \Rightarrow A< \frac{1}{2}\)
Chúc bạn học tốt!
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
đề cần chứng minh nhỏ hơn 1 hay 11
nếu 1 thì
\(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{100^2}\)
\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.......+\frac{1}{99\cdot100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrowđcm\)
nếu nhỏ hơn 11 thì làm như thế thêm câu
vì đẳng thức trên <1<11
=>đcm
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\frac{2}{x+1}=1-\frac{2007}{2009}\)
\(\frac{2}{x+1}=\frac{2}{2009}\)
\(\Rightarrow x+1=2009\Leftrightarrow x=2008\)
_ giải bừa :v _
\(T=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}\)
Ta thấy : \(\frac{1}{4^2}< \frac{1}{2.4};\frac{1}{14^2}< \frac{1}{12.14}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{14^2}< \frac{1}{2^2}+\frac{1}{2.4}+...+\frac{1}{12.14}\)
\(\Rightarrow T< \frac{1}{2^2}+\frac{1}{2}\left(\frac{2}{2.4}+...+\frac{2}{12.14}\right)\)
\(\Rightarrow T< \frac{1}{2^2}+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(\Rightarrow T< \frac{1}{4}+\frac{1}{2}.\frac{3}{7}\)
\(\Rightarrow T< \frac{13}{28}\)
Mà \(\frac{13}{28}< \frac{1}{2}\Rightarrow T< \frac{1}{2}\)
....
Ta có:
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(\frac{1}{5^2}< \frac{1}{4.5}\)
....
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}\)
\(-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
=> đpcm
Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\left(đpcm\right)\)
Ta có: \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)