CMR 1/2+1/2^2+...+1/2^100<1
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Ta có :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=1.100-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}\right)\)
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+.......+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+.........+\frac{99}{100}\)
Vậy \(100-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+.....+\frac{99}{100}\left(ĐPCM\right)\)
\(=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\Rightarrow A< 1\)
A=122+132+142+...+11002�=122+132+142+...+11002
A<11⋅2+12⋅3+13⋅4+...+199⋅100�<11⋅2+12⋅3+13⋅4+...+199⋅100
A<11−12+12−13+13−14+...+199−1100�<11−12+12−13+13−14+...+199−1100
A<1−1100�<1−1100
A<99100�<99100
Mà 99100<1⇒A<1
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
\(\Rightarrow2A-A=1-\dfrac{1}{2^{100}}\)
\(\Rightarrow A=1-\dfrac{1}{2^{100}}< 1\) (đpcm)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{100}}\)
\(2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}\)
\(2A-A=1-\dfrac{1}{2^{100}}\)
\(A=1-\dfrac{1}{2^{100}}< 1\left(đpcm\right)\)