CMR:\(\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}-\frac{100}{2^{100}}<\frac{2}{9}\)
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SY
18 tháng 3 2016
Đặt A=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) +\(\frac{3}{3^3}\) - \(\frac{4}{3^4}\)+...+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3A=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\)+...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4A=1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\)- \(\frac{100}{3^{100}}\)
=> 4A<1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\) (1)
Đặt B=1-\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{3^{98}}\) - \(\frac{1}{3^{99}}\)
=> B=2+ \(\frac{1}{3}\) - \(\frac{1}{3^2}\) +...+\(\frac{1}{3^{97}}\) - \(\frac{1}{3^{98}}\)
=> 4B=B+3B=3-\(\frac{1}{3^{99}}\)<3 => A<\(\frac{3}{4}\) (2)
Từ (1) và (2) ta có: 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
27 tháng 6 2020
Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+....\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....\frac{1}{3^{98}}-\frac{1}{3^{99}}\Rightarrow4A< B\left(1\right)\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+....\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\Rightarrow4B< 3\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4A< B< \frac{3}{4}\Rightarrow4A< \frac{3}{4}\Rightarrow A< \frac{3}{4}:4\Rightarrow A< \frac{3}{4}.\frac{1}{4}\Rightarrow A< \frac{3}{16}\)
=> đpcm.
14 tháng 1 2018
Có : (1+1/2+1/3+....+1/100)+(1/2+2/3+....+99/100)
= 1+(1/2+1/2)+(1/3+2/3)+.....+(1/100+99/100) ( có 99 cặp )
= 1+1+1+....+1 ( có 100 số 1 )
= 100
=> 100-(1+1/2+1/3+....+1/100)=1/2+2/3+3/4+....+99/100
Tk mk nha
Lời giải:
$A=\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}-....+\frac{99}{2^{99}}-\frac{100}{2^{100}}$
$2A=1-\frac{2}{2}+\frac{3}{2^2}-....+\frac{99}{2^{98}}-\frac{100}{2^{99}}$
$\Rightarrow A+2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...-\frac{1}{2^{99}}-\frac{100}{2^{100}}$
$\Rightarrow 3A+\frac{100}{2^{100}}=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+...-\frac{1}{2^{99}}$
$\Rightarrow 2(3A+\frac{100}{2^{100}}) =2-1+\frac{1}{2}-\frac{1}{2^2}+...-\frac{1}{2^{98}}$
$\Rightarrow 3A+\frac{100}{2^{100}}+2(3A+\frac{100}{2^{100}})=2-\frac{1}{2^{99}}$
$\Rightarrow 9A+\frac{300}{2^{100}}=2-\frac{1}{2^{99}}$
$\Rightarrow 9A=2-\frac{1}{2^{99}}-\frac{300}{2^{100}}<2$
$\Rightarrow A< \frac{2}{9}$