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\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)Mà \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}< 1\)\(\Rightarrow A< 1\)
Đặt : \(B=\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
\(B=\left(\dfrac{99}{1}+1\right)+\left(\dfrac{98}{2}+1\right)+...+\left(\dfrac{1}{99}+1\right)-99\)
\(B=\dfrac{100}{1}+\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}-99\)
\(B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\left(100-99\right)\)
\(B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\dfrac{100}{100}\)
\(B=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)
Ta có : \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
\(\Rightarrow2^{100}.A=2^{99}-2^{98}+2^{97}-2^{96}+...+2-1\)
\(\Rightarrow2^{101}.A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2^{100}.A+2^{101}.A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2+\left(2^{99}-2^{98}+2^{97}-2^{96}+...+2-1\right)\)
\(\Rightarrow A\left(2^{100}+2^{101}\right)=2^{100}-1\)
\(\Rightarrow A=\dfrac{2^{100}-1}{2^{100}+2^{101}}\)
sửa đề : \(F=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(\dfrac{1}{1^2}< \dfrac{1}{1.2};\dfrac{1}{2^2}< \dfrac{1}{2.3};...;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
Cộng vế với vế
\(\dfrac{1}{1^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)< 7/4
Vậy ta có đpcm
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
\(\Rightarrow2A=\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{1}{2^{101}}\)
\(\Rightarrow3A=2A+A\)
\(=\left(\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{1}{2^{101}}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)
\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{2^{101}}\right):3\)
\(=\dfrac{1}{6}-\dfrac{1}{3.2^{101}}\)
o cho 1/2 mu 4 +... thi la + hay - ?
minh thay hoi la