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LG
1
DH
18 tháng 2 2020
Đặt: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{2019}{3^{2018}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2018}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2017}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2018}}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^{2018}}}{2}\)
Thay vào \(2A\Rightarrow2A=1+\frac{\left(1-\frac{1}{3^{2018}}\right)}{2}-\frac{2019}{3^{2019}}\)
\(=1+\frac{1}{2}-\frac{1}{2.3^{2018}}-\frac{2019}{3^{2019}}< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< 0,75\left(đpcm\right)\)
OT
1
15 tháng 2 2020
Đặt A=\(\frac{1}{3}+\frac{2}{3^2}+.....+\frac{2019}{3^{2019}}\)
3A=\(1+\frac{2}{3}+.....+\frac{2019}{3^{2018}}\)
3A - A = \(\left(1+\frac{2}{3}+...+\frac{2018}{3^{2017}}+\frac{2019}{3^{2018}}\right)\) -\(\left(\frac{1}{3}+....+\frac{2017}{3^{2017}}+\frac{2018}{3^{2018}}+\frac{2019}{3^{2019}}\right)\)
2A = \(1+\frac{1}{3}+...+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt B=\(1+\frac{1}{3}+....+\frac{1}{3^{2018}}\)
3B =\(3+1+....+\frac{1}{3^{2017}}\)
3B - B=\(\left(3+1+....+\frac{1}{3^{2017}}\right)\)-\(\left(1+\frac{1}{3}+...+\frac{1}{3^{2018}}\right)\)
2B =\(3-\frac{1}{3^{2018}}\)
Ta có:2A= B - \(\frac{2019}{3^{2019}}\)
4A = 2B -\(\frac{2.2019}{3^{2019}}\)
4A=\(\left(3-\frac{1}{3^{2018}}\right)\)-\(\frac{2.2019}{3^{2019}}\)
A=\(\frac{3}{4}-\frac{1}{3^{2018}.4}-\frac{2019}{3^{2019}.2}\)<\(\frac{3}{4}\)=0,75
Suy ra :\(\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)< 0,75 (đpcm)
7 tháng 2 2020
Ta có:
\(a=1-\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2-\left(\frac{2019}{2020}\right)^3+...+\left(\frac{2019}{2020}\right)^{2020}\)
=> \(\frac{2019}{2020}.a=\frac{2019}{2020}-\left(\frac{2019}{2020}\right)^2+\left(\frac{2019}{2020}\right)^3-...+\left(\frac{2019}{2020}\right)^{2020}-\left(\frac{2019}{2020}\right)^{2021}\)
Lấy
\(a+\frac{2019}{2020}a=1-\left(\frac{2019}{2020}\right)^{2021}\)
<=> \(a\left(1+\frac{2019}{2020}\right)=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)
<=> \(a.\frac{4039}{2020}=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)
<=> \(a.=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}\)
Vì : \(0< \left(\frac{2019}{2020}\right)^{2021}< 1\)
=> \(0< 1-\left(\frac{2019}{2020}\right)^{2021}< 1\)
và \(0< \frac{2020}{4039}< 1\)
=> \(0< \left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}< 1\)
=> 0 < a < 1
=> a không phải là một số nguyên.
T
0
6 tháng 1 2023
\(\dfrac{2}{3}A=\dfrac{2}{3}-\left(\dfrac{2}{3}\right)^2+\left(\dfrac{2}{3}\right)^3-...+\left(\dfrac{2}{3}\right)^{2019}-\left(\dfrac{2}{3}\right)^{2020}\)
=>\(\dfrac{5}{3}A=1-\left(\dfrac{2}{3}\right)^{2020}=1-\dfrac{2^{2020}}{3^{2020}}=\dfrac{3^{2020}-2^{2020}}{3^{2020}}\)
=>\(A=\dfrac{3^{2020}-2^{2020}}{3^{2020}}:\dfrac{5}{3}=\dfrac{3^{2020}-2^{2020}}{5\cdot3^{2019}}\) ko là số nguyên
JG
0
ko bit