\(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
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14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
18 tháng 2 2022
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
25 tháng 2 2022
\(S=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\\ 2S=1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{1}{2^{99}}\\ 2S+S=1-\dfrac{1}{2^{100}}\\ S=\dfrac{1-\dfrac{1}{2^{100}}}{3}\)
KV
22 tháng 3 2023
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
\(\Rightarrow2A=\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{1}{2^{101}}\)
\(\Rightarrow3A=2A+A\)
\(=\left(\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{100}}-\dfrac{1}{2^{101}}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{2^{101}}\)
\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{2^{101}}\right):3\)
\(=\dfrac{1}{6}-\dfrac{1}{3.2^{101}}\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
\(\Rightarrow2^{100}.A=2^{99}-2^{98}+2^{97}-2^{96}+...+2-1\)
\(\Rightarrow2^{101}.A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow2^{100}.A+2^{101}.A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2+\left(2^{99}-2^{98}+2^{97}-2^{96}+...+2-1\right)\)
\(\Rightarrow A\left(2^{100}+2^{101}\right)=2^{100}-1\)
\(\Rightarrow A=\dfrac{2^{100}-1}{2^{100}+2^{101}}\)
= 1 phần 2 bạn nhé