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Vì mấy cái kia cộng lại sẽ bằng 0,477..
1/2=0,5 nên tổng kia sẽ nhỏ hơn 1/2
Ta có\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}< \frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}\)(A là đề bài)
Mà \(\frac{1}{5}-\frac{1}{30}=\frac{1}{6}< \frac{1}{5}-\frac{1}{101}< A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{6}< A< \frac{1}{4}\left(ĐPCM\right)\)
Ta có: \(\frac{1}{5\cdot6}< \frac{1}{5^2}=\frac{1}{5\cdot5}< \frac{1}{4\cdot5}\)
\(\frac{1}{6\cdot7}< \frac{1}{6^2}=\frac{1}{6\cdot6}< \frac{1}{5\cdot6}\)
\(\frac{1}{7\cdot8}< \frac{1}{7^2}=\frac{1}{7\cdot7}< \frac{1}{6\cdot7}\)
.............................
\(\frac{1}{100\cdot101}< \frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
Đặt \(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{100\cdot101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\)
\(B=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(=>\frac{1}{6}< A< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< B< \frac{1}{4}\)
\(=>\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(Đpcm\right)\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}< 1\)
Vậy \(A< 1\left(đpcm\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}-\frac{1}{50}\)
\(\Leftrightarrow B< \frac{1}{4}+\frac{1}{2}\)
\(\Leftrightarrow B< \frac{3}{4}\left(đpcm\right)\)
chịu thui
chúc bn học gioi!
nhaE@@
Toán lớp 7 bye mk đi hc đây hihi
$$$$
Ta có : \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
= \(\frac{1}{3}+\)( \(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\)) \(+\)( \(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\) ) \(< \)\(\frac{1}{3}+\)( \(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\)) \(+\)( \(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\)) = \(\frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\)