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\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
=> \(A=1-\frac{1}{2^{99}}<1\)
=> \(A<1\)(Đpcm)
Ta có:
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\)
\(2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}<1\left(đpcm\right)\)
\(A=\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+...........+\frac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{99}}\)
\(\Leftrightarrow2^{99}.A=2^{99}-1\left(đpcm\right)\)
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)
\(B=1-\frac{1}{2^{99}}< 1\left(đpcm\right)\)
B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
=> 2B=\(2\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]\)
=\(1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{98}\)
=>2B-B=\(\left[1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{98}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\right]\)
=>B=\(1-\left(\frac{1}{2}\right)^{99}< 1\)
=> B<1
\(B=\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+........+\frac{1^{99}}{2^{99}}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{99}}\right)\)
=>B=\(1-\frac{1}{2^{98}}\Rightarrow B<1\)
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)
\(A=1-\frac{1}{2^{99}}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
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