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Ta có :
\(\frac{1}{5^2}>\frac{1}{5.6}\)
\(\frac{1}{6^2}>\frac{1}{6.7}\)
\(..............\)
\(\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A>\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\left(1\right)\)
Lại có :
\(\frac{1}{5^2}< \frac{1}{4.5}\)
\(\frac{1}{6^2}< \frac{1}{5.6}\)
\(...............\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)
Từ (1) và (2) => Điều phải chứng minh
Đặt S = \(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\)
=> 72S = 49S = \(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\)
=> 49S - S = \(\left(1+\frac{1}{7^2}+\frac{1}{7^4}+...+\frac{1}{7^{98}}\right)-\left(\frac{1}{7^2}+\frac{1}{7^4}+\frac{1}{7^6}+...+\frac{1}{7^{100}}\right)\)
=> 48S = \(1-\frac{1}{7^{100}}\)
=> \(S=\frac{1-\frac{1}{7^{100}}}{48}\)
Khi đó A = \(\left(\frac{1-\frac{1}{7^{100}}}{48}\right):\left(1-\frac{1}{7^{100}}\right)=\frac{1}{48}\)
Ta có: \(\frac{1}{5^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{100^2}< \frac{1}{99.100}\)
Cộng vế với vế ta được: \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)(1)
Tương tự: \(\frac{1}{5^2}>\frac{1}{5.6};\frac{1}{6^2}>\frac{1}{6.7};...;\frac{1}{100^2}>\frac{1}{100.101}\)
Cộng vế với vế ta được \(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\)(2)
Từ (1) và (2) =>đpcm
Đặt S = \(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\)
=> 6S = \(1+\frac{1}{6}+\frac{1}{6^2}+...+\frac{1}{6^{99}}\)
=> 6S - S = \(\left(1+\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{99}}\right)-\left(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\right)\)
=> \(5S=1-\frac{1}{6^{100}}\)
=> \(S=\frac{1-\frac{1}{6^{100}}}{5}\)
Khi đó A = \(\left(1-\frac{1}{6^{100}}\right):\left(\frac{1-\frac{1}{6^{100}}}{5}\right)=5\)
\(Ta\) \(có : \)
\(1 / 5^2 + 1 /6^2 + ... + 1 /100^2 < 1 /4.5\)\(+ 1 / 5 .6 + ... + 1 / 99 .100\)
\(Mà ta có:\)\(1 / 4 .5 + 1 / 5 .6 + ... + 1 / 99 .100\)
\(\Rightarrow\)\(1 / 4 - 1 / 5 + 1 / 5 - 1 / 6 + ... +\)\(1 / 99 - 1 / 100\)
\(\Rightarrow\)\(1 / 4 - 1 / 100\) \(< 1 / 4\)
\(Nên 1 / 5^2 + 1 /6^2 + ...+ 1 / 100^2 < 1 / 4\)
Tương tự chứng minh tiếp nhé 😘😘
C/m<1/4
t\(n^2>n\left(n-1\right)=>\frac{1}{n^2}<\frac{1}{n\left(n-1\right)}\)
\(\frac{1}{5^2}<\frac{1}{4.5};\frac{1}{6^2}<\frac{1}{5.6};\frac{1}{100^2}<\frac{1}{99.100}\)
\(\frac{1}{4.5}+..+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}<\frac{1}{4}.ok\)
CM>1/6
\(n^2\frac{1}{n\left(n+1\right)}\)
\(\frac{1}{5.6}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{5}>\frac{1}{6}\)OK
doan cuoi
\(\frac{1}{5}-\frac{1}{101}=\frac{96}{5.101}>\frac{96}{5.102}=\frac{1.}{6}.\frac{96}{85}>\frac{1}{6}ok\)