So sánh
\(\dfrac{1}{5^1}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+....+\dfrac{1}{5^{2015}}\)
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21 tháng 7 2017
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}< \dfrac{1}{4}\)
\(\Rightarrowđpcm\)
21 tháng 7 2017
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\)
\(\Rightarrow5A=5\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow5A=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\)
\(\Rightarrow5A-A=\left(1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\)
\(\Rightarrow4A=1-\dfrac{1}{5^{2015}}\)
\(\Rightarrow A=\dfrac{1}{4}-\dfrac{1}{5^{2015}.4}\)
\(\Rightarrow A< \dfrac{1}{4}\)
28 tháng 2 2021
Bạn thiếu đề rồi phải là trừ hay cộng j j chứ.
Xét:
`A+B=2+1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025`
`1/2+1/3+1/4+......+1/4026+1/3+1/5+1/7+......+1/4025>0`
`=>A+B>2`
Mà `1 2013/2014<2`
`=>A+B>1 2013/2014`
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) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
14 tháng 1 2023
Bài 1:
a: Sửa đề: 1/3^200
1/2^300=(1/8)^100
1/3^200=(1/9)^100
mà 1/8>1/9
nên 1/2^300>1/3^200
b: 1/5^199>1/5^200=1/25^100
1/3^300=1/27^100
mà 25^100<27^100
nên 1/5^199>1/3^300
AH
Akai Haruma
Giáo viên
15 tháng 7 2023
Lời giải:
$M=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}$
$5M=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}$
$\Rightarrow 4M=5M-M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}$
$4M+\frac{2014}{5^{2014}}=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}$
$5(4M+\frac{2014}{5^{2014}})=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}$
$\Rightarrow 4(4M+\frac{2014}{5^{2014}})=5-\frac{1}{5^{2013}}$
$M=\frac{5}{16}-\frac{1}{16.5^{2013}-\frac{2014}{4.5^{2014}}$
16 tháng 2 2022
\(\dfrac{1}{4444}< 1,\dfrac{3}{7}< 1,\dfrac{9}{5}>1,\dfrac{7}{3}>1,\dfrac{14}{15}< 1,\dfrac{16}{16}=1,\dfrac{14}{11}>1\)
\(A=\dfrac{1}{5^1}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\\ 5A=1+\dfrac{1}{5^1}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\\ 5A-A=\left(1+\dfrac{1}{5^1}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2013}}+\dfrac{1}{5^{2014}}\right)-\left(\dfrac{1}{5^1}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{2014}}+\dfrac{1}{5^{2015}}\right)\\ 4A=1-\dfrac{1}{5^{2015}}\Rightarrow A=\dfrac{1-\dfrac{1}{5^{2015}}}{4}=\dfrac{1}{4}-\dfrac{4}{5^{2015}}< \dfrac{1}{4}\)