c/m: 1/3+2/3*2+3/3*3....+2019/3*2019 bé hơn 0,75
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15 tháng 4 2019
Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+...+\frac{2019}{3^{2019}}\)
=>\(3A=1+\frac{2}{3}+...+\frac{2019}{3^{2018}}\)
=>\(2A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2018}}-\frac{2019}{3^{2019}}\)
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{2018}}\)
=>\(2B=3-\frac{1}{3^{2018}}\)=>\(B=\frac{3-\frac{1}{3^{2018}}}{2}\)
=>\(2A=\frac{3-\frac{1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}=\frac{\frac{3^{2019}-1}{3^{2018}}}{2}-\frac{2019}{3^{2019}}\)
\(=\frac{3^{2019}-1}{3^{2018}.2}-\frac{2019}{3^{2019}}=\frac{3\left(3^{2019}-1\right)-2019.2}{3^{2019}.2}\)
15 tháng 4 2019
Nhầm tí
dòng thứ 2 từ dưới lên cm bé hơn 0,75 luôn nhá
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)
T
0
BC
16 tháng 1 2019
Ta có:
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{2019}}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{2018}}\)
\(\Rightarrow3A-A=\left(1+\frac{1}{3}+....+\frac{1}{3^{2018}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{2019}}\right)\)
\(\Rightarrow2A=1+\frac{1}{3}+....+\frac{1}{3^{2018}}-\frac{1}{3}-\frac{1}{3^2}-.....-\frac{1}{3^{2019}}\)
\(\Rightarrow2A=1-\frac{1}{3^{2019}}\)
\(\Rightarrow\)\(A=\left(1-\frac{1}{3^{2019}}\right):2\)
\(\Rightarrow A=1:2-\frac{1}{3^{2019}}:2\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3^{2019}}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy \(A< \frac{1}{2}\)