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Ta có : \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
Nên \(A< B\)
\(\Rightarrow A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)
\(\Rightarrow A.B=\frac{1}{201}\)
Vì \(A< B\)
\(\Rightarrow A^2< A.B=\frac{1}{201}\)
\(\Rightarrow A^2< \frac{1}{201}\)
\(\RightarrowĐPCM\)
Ta có \(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
.....................
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + ... + 1/100^2 < 1/2
1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + ... + 1/100.100 < 1/2.3+ 1/3.4 + 1/4 .5 + 1/5.6 + .. + 1/99.100
1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + ... + 1/100.100 < 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/99 - 1/100
1/3.3 + 1/4.4 + 1/5.5 + 1/6.6 + ... + 1/100.100 < 1/2 - 1/100 suy ra 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + ... + 1/100^2 < 1/2
Chúc bn hok tốt
M = \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\)
=> 5M = 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)
=> 5M - M = ( 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)) - ( \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\))
4M = 1 - \(\left(\frac{1}{5}\right)^{50}\)
=> M = \(\frac{1-\left(\frac{1}{5}\right)^{50}}{4}\)< \(\frac{1}{4}\)
Ta có:
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
\(...\)
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\) \(\left(1\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
Đặt \(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{3}-\dfrac{1}{100}\)\(< \dfrac{1}{3}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)