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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
a) gọi \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
gọi \(B=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
\(\Rightarrow A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\)
b) Ta thấy \(\frac{1}{37}< \frac{1}{35}< \frac{1}{31}< \frac{1}{30}\), \(\frac{1}{61}< \frac{1}{53}< \frac{1}{47}< \frac{1}{45}\)
Do đó : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{53}+\frac{1}{61}< \frac{1}{3}+\frac{1}{30}.3+\frac{1}{45}.3=\frac{1}{2}\)
c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{2500}\right)\)
\(=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Ta thấy vế trong ngoặc nhỏ hơn 1
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>48\)
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 : \(\frac{2}{5}< A< \frac{8}{9}\)
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(\dfrac{1}{7^2}< \dfrac{1}{6.7}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
A<\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
A<\(\left[\left(\dfrac{1}{1}-\dfrac{1}{8}\right)+\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(-\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(-\dfrac{1}{5}+\dfrac{1}{5}\right)+\left(-\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{7}+\dfrac{1}{7}\right)+\left(-\dfrac{1}{8}+\dfrac{1}{8}\right)\right]\)A<\(\left[\left(\dfrac{8}{8}-\dfrac{1}{8}\right)+0+0+0+0+0+0+0\right]\)
A<\(\dfrac{7}{8}< 1\)
Vậy ta có đpcm.
Sorry nha, chỗ phân tích ra thành tống đại số phải như này :
A<\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
Rồi làm tương tự.
Mình cho bạn công thức nè :\(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)