A =1/2 + 1/2^2 + 1/2^3 + .. + 1/2^100. CMR A < 1
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11 tháng 3 2022
\(A = \dfrac{1}{2^2} + \dfrac{1}{4^2} +\dfrac{1}{6^2} +...... +\dfrac{1}{100^2} \)
\(A = \dfrac{1}{1^2.2^2} +\dfrac{1}{2^2.2^2} +\dfrac{1}{2^2.3^2} + .......+\dfrac{1}{2^2.2^{50}}\)
\(A = \dfrac{1}{2^2}.(\) \( \dfrac{1}{1^2} + \dfrac{1}{2^2} +\dfrac{1}{3^2} +...... +\dfrac{1}{50^2}) \)
\(A < \dfrac{1}{2^2}.( \dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{49.50}\) \()\)
\(= \dfrac{1}{2^2}.(1-\dfrac{1}{2} + \dfrac{1}{2}-\dfrac{1}{3}+.......+\dfrac{1}{49}-\dfrac{1}{50})\)
\(= \dfrac{1}{2^2} . ( 1 - \dfrac{1}{50})\)
\(< \dfrac{1}{2^2} . 2 = \dfrac{1}{2}\)
1 tháng 4 2023
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(A< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< 1-\dfrac{1}{100}\)
\(A< \dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\Rightarrow A< 1\)
1 tháng 4 2023
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
TN
25 tháng 5 2016
\(2A=2\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{100}}\)
Đến đây tôi chịu
25 tháng 5 2016
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=A=\frac{1}{2}-\frac{1}{2^{100}}\)
\(A+\frac{1}{2^{100}}=\frac{1}{2}-\frac{1}{2^{100}}+\frac{1}{2^{100}}=\frac{1}{2}\)
Vậy \(A+\frac{1}{2^{100}}=\frac{1}{2}\)
NM
1
24 tháng 4 2015
Ta thấy:\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
27 tháng 9 2023
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{99.100}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
9 tháng 2 2017
TA CÓ 1/2^2=1/4
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
1/100^2<1/99.100
=>1/2^2+2/3^2+.....+1/100^2<1/1.2+1/2.3+..+1/99.100
=1-99/100=99/100<1
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^{100}}\)
\(2A-A=\left(\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(A=\frac{2}{2}+\frac{2}{2^2}+..+\frac{2}{2^{100}}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{100}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-\frac{1}{2}-...-\frac{1}{2^{100}}\)
\(A=1-\frac{1}{2^{100}}\)
\(\Rightarrow A< 1\)