\(A=\dfrac{1}{^{ }3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\) và \(B=\dfrac{1}{2}\). Hãy so sánh chúng
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28 tháng 6 2021
Ta có `3A=1+1/3+....+1/3^99`
`=>3A-A=1-1/3^100`
`=>2A=1-1/3^100`
`=>A=1/2-1/(2.3^100)<1/2`
Hay `A<B`
17 tháng 8 2021
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
AH
Akai Haruma
Giáo viên
22 tháng 5 2023
Lời giải:
$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2022}}$
$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2021}}$
$\Rightarrow 3A-A=1-\frac{1}{3^{2022}}$
$\Rightarrow A=\frac{1}{2}-\frac{1}{2.3^{2022}}$
Xét hiệu:
$A-B=\frac{1}{2}-\frac{1}{2.3^{2022}}-(1-\frac{1}{3^{2021}})$
$=\frac{1}{3^{2021}}-\frac{1}{2.3^{2022}}-\frac{1}{2}$
$=\frac{5}{2.3^{2022}}-\frac{1}{2}$
$< \frac{1}{2}-\frac{1}{2}=0$
$\Rightarrow A< B$
8 tháng 7 2023
\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)
\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)
\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)
\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)
\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)
\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)
mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)
\(\Rightarrow A< B\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 7 2023
A = \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)
3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)
3A - A = 1 - \(\dfrac{1}{3^{2023}}\)
2A = 1 - \(\dfrac{1}{3^{2023}}\) < 1
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)
B = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)
B = \(\dfrac{8}{12}\)
B = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1
2A < 2B ⇒ A < B
23 tháng 3 2023
\(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{99}}\)
=>2S=1-1/3^100
=>S=1/2-1/2*3^100<1/2
MH
6 tháng 5 2022
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
Ta có: \(3\cdot A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
Do đó:
\(3\cdot A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)
hay \(2\cdot A=1-\dfrac{1}{3^{100}}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right):2\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right)\cdot\dfrac{1}{2}\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{100}}< \dfrac{1}{2}\)
hay A<B
Ta có: 3⋅A=1+131+132+...+13993⋅A=1+131+132+...+1399
A=13+132+...+13100A=13+132+...+13100
Do đó:
3⋅A−A=1+131+132+...+13100−13−132−...−131003⋅A−A=1+131+132+...+13100−13−132−...−13100
hay 2⋅A=1−131002⋅A=1−13100
⇔A=(1−13100):2⇔A=(1−13100):2
⇔A=(1−13100)⋅12⇔A=(1−13100)⋅12
⇔A=12−12⋅3100<12⇔A=12−12⋅3100<12
hay A<B