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Ta có:B = \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}=\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+1-\frac{1}{2^{100}}\)
\(=1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}=100-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
=> \(B=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\) (Đpcm)
Ta có \(\frac{1}{2}+\frac{3}{2^2}+\frac{7}{2^3}+...+\frac{2^{100}-1}{2^{100}}\)
= \(\frac{2-1}{2}+\frac{2^2-1}{2^2}+\frac{2^3-1}{2^3}+...+\frac{2^{100}-1}{2^{100}}\)
= \(1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^{100}}\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
\(=100-\left(1-\frac{1}{2^{100}}\right)=100-1+\frac{1}{2^{100}}=99+\frac{1}{2^{100}}>99\)(đpcm)
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)
Đặt S = \(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\)
=> 6S = \(1+\frac{1}{6}+\frac{1}{6^2}+...+\frac{1}{6^{99}}\)
=> 6S - S = \(\left(1+\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{99}}\right)-\left(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+...+\frac{1}{6^{100}}\right)\)
=> \(5S=1-\frac{1}{6^{100}}\)
=> \(S=\frac{1-\frac{1}{6^{100}}}{5}\)
Khi đó A = \(\left(1-\frac{1}{6^{100}}\right):\left(\frac{1-\frac{1}{6^{100}}}{5}\right)=5\)
\(A=\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\)
\(\frac{1}{5^2}A=\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+...+\frac{1}{5^{103}}\)
\(\left(1-\frac{1}{5^2}\right)A=\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)-\left(\frac{1}{5^3}+\frac{1}{5^5}+\frac{1}{5^7}+...+\frac{1}{5^{103}}\right)\)
\(\frac{24}{25}A=\frac{1}{5}-\frac{1}{5^{103}}\)
\(A=\left(1-\frac{1}{5^{102}}\right).\frac{5}{24}\)
Suy ra \(\left(\frac{1}{5}+\frac{1}{5^3}+\frac{1}{5^5}+...+\frac{1}{5^{101}}\right)\div\left(1-\frac{1}{5^{102}}\right)=\frac{5}{24}\).
Sửa đề \(\frac{3}{2}+\frac{5}{2^2}+\frac{9}{2^3}+...+\frac{2^{100}+1}{2^{100}}=\frac{2+1}{2}+\frac{2^2+1}{2^2}+\frac{2^3+1}{2^3}+...+\frac{2^{100}+1}{2^{100}}\)
= \(\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)(100 hạng tử 1)
= \(100+\left(1-\frac{1}{2^{100}}\right)=101-\frac{1}{2^{100}}< 101\)(1)
Vì \(-\frac{1}{2^{100}}>-1\Rightarrow101-\frac{1}{2^{100}}>101-1\Rightarrow B>100\)(2)
Từ (1) và (2) => 100 < B < 101