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\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2015}\)
\(B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(2B=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(B=1-\frac{1}{2^{2015}}< 1\). Vậy ta có điều phải chứng minh
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
=> \(A=1-\frac{1}{2^{99}}<1\)
=> \(A<1\)(Đpcm)
Ta có:
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\)
\(2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{99}}\)
\(A=1-\frac{1}{2^{99}}<1\left(đpcm\right)\)
\(B=\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+........+\frac{1^{99}}{2^{99}}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{99}}\right)\)
=>B=\(1-\frac{1}{2^{98}}\Rightarrow B<1\)
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{2014}\)
\(A=\frac{1}{2}+\frac{1^2}{2^2}+...+\frac{1^{2014}}{2^{2014}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\)
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{2013}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{2013}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}\right)\)
\(A=1-\frac{1}{2^{2014}}< 1\)
Đpcm