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Phần a, A> 1/3.4+1/4.5+1/5.6+...+ 1/50.51 = 1/3-1/4+1/4-1/5+1/5-1/6+...+ 1/50-1/51 = 1/3-1/51 = 48/153 > 48/192 =1/4. ĐPCM
Phần b, A< 1/3^2+1/3.4+1/4.5+...+1/49.50 = 1/9+1/3-1/4+1/4-1/5+...+ 1/49-1/50 = 1/9+1/3-1/50 = 1/9+47/150 < 1/9+50/150 = 1/9+1/3 = 4/9. ĐPCM
Ta có
\(A>\frac{1}{3^2}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}\)
\(\Rightarrow A>\frac{1}{9}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{50}-\frac{1}{51}\)
\(\Rightarrow A>\frac{1}{4}+\left(\frac{1}{9}-\frac{1}{51}\right)\)
\(\Rightarrow A>\frac{1}{4}+\frac{42}{9.51}>\frac{1}{4}\)
Vậy A>1/4
b)
Ta có
\(A< \frac{1}{3}^2+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{9}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{50}\)
\(\Rightarrow A< \frac{4}{9}-\frac{1}{50}< \frac{4}{9}\)
Vậy A<4/9
Ta có :
\(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};...;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{5}{10}-\frac{1}{10}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{4}{10}=\frac{2}{5}\left(1\right)\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< 1-\frac{1}{9}=\frac{8}{9}\left(2\right)\)
Từ ( 1 ) , ( 2 ) => ĐPCM
Chúc bạn học tốt !!!
Đề sai bạn nhé :
Đề đúng :
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
CM : \(\frac{2}{5}< A< \frac{8}{9}\)
2. \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2}{9}\)
\(\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{2}{9}\)
\(2.\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{2}{9}\)
\(2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)
\(2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}\right)\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{2}{9}:2\)
\(\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)
\(\dfrac{1}{x+1}=\dfrac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
ta có A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
= \(1-\frac{1}{9}\)
= \(\frac{8}{9}\)
suy ra A < \(\frac{8}{9}\)
ta có A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)j> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
= \(\frac{1}{2}-\frac{1}{10}\)
= \(\frac{2}{5}\)
suy ra A >\(\frac{2}{5}\)
a) Ta thấy: 1/2^2<1/1.2
1/3^2<1/2.3
1/4^2<1/3.4
…………...
1/100^2<1/99.100
=>A<1/1.2+1/2.3+1/3.4+…+1/99.100=99/100
Mà 99/100<1 => 1/22 + 1/32 + 1/42 + ... + 1/1002<1
b)Ta thấy : 1/101+1/102+1/103+…+1/150>1/150+1/150+1/150+…+1/150(50 số hạng)
=>A>50/150>1/3 (1)
Ta thấy : 1/101+1/102+1/103+…+1/150<1/100+1/100+1/100+…+1/100(50 số hạng)
=>A<1/2 (2)
Từ (1) và (2) =>1/3<A<1/2
c) Ta thấy : 1/11 + 1/12 + 1/13 + ... + 1/20>1/20+1/20+1/20+…+1/20(10 số hạng)
=>1/11 + 1/12 + 1/13 + ... + 1/20>1/2