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Ta có 4A=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
Trừ 4A cho A ta được
3A = \(1-\frac{1}{2^{100}}\)=> 3A <1 => A<1/3 (đpcm)
Chúc bạn học tốt
Ta có :\(A=\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(2A=\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{100}}\)
Lại có :
\(\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)
Vì \(\frac{1}{2^{100}}< \frac{1}{6}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2^{100}}>\frac{1}{2}-\frac{1}{6}\)
\(\Rightarrow A>\frac{1}{3}\)
Vậy \(A>\frac{1}{3}\)(ĐPCM)
Ta có : \(A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+......+\frac{1}{2^{100}}\)
\(\Rightarrow4A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^4}+.....+\frac{1}{2^{98}}\)
\(\Rightarrow4A-A=\frac{1}{2}-\frac{1}{2^{100}}\)
\(\Rightarrow3A=\frac{2^{99}-1}{2^{100}}\)
\(\Rightarrow A=\frac{2^{99}-1}{\frac{2^{200}}{3}}\)
Vì : \(\frac{2^{99}-1}{2^{200}}< 1\)
Nên : \(A< \frac{1}{3}\)
Ta thấy : \(\frac{1}{2^2}< \frac{1}{3}\)
\(\frac{1}{2^4}< \frac{1}{3}\)
...
\(\frac{1}{2^{100}}< \frac{1}{3}\)
\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{2^4}+\frac{1}{2^6}+...+\frac{1}{2^{100}}< \frac{1}{3}\)
Vậy \(A< \frac{1}{3}\)
Chúc bạn học tốt :>
A.\(4\)=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)
=> 4A-A=1-\(\frac{1}{2^{100}}\)
=> A=\(\frac{1}{3}\left(1-\frac{1}{2^{100}}\right)=\frac{1}{3}-\frac{1}{3}.\frac{1}{2^{100}}< \frac{1}{3}\)
4A = 1 +1/2^2+1/2^4+....+1/2^98
3A = 4A-A = (1+1/2^2+1/2^4+....+1/2^98) - (1/2^2+1/2^4+....+1/2^100) = 1 - 1/2^100 < 1
=> A < 1/3 ( ĐPCM )
k mk nha
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(A< \frac{1}{4}.\left(2-\frac{1}{50}\right)< \frac{1}{4}.2=2\)
=> \(A< 2\left(đpcm\right)\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}.\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
\(A< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\right)\)
\(A< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(A< \frac{1}{4}.\left(2-\frac{1}{50}\right)< \frac{1}{4}.2=2\)
\(A< 2\left(đpcm\right)\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow A< 1-\frac{1}{9}=\frac{8}{9}\)(1)
Lại có: \(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(2)
Từ (1) và (2), suy ra: \(\frac{2}{5}< A< \frac{8}{9}\)
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