Cho A=\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{2019}{3^{504}}\)
Chứng tỏ A<\(\frac{9}{2}\)
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Những câu hỏi liên quan
NT
0
HM
2
4 tháng 4 2019
Đặt A=\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+\frac{19}{3^4}+...+\frac{2015}{3^{503}}+\frac{2019}{3^{504}}\)
3A=\(7+\frac{11}{3}+\frac{15}{3^2}+\frac{19}{3^3}+...+\frac{2015}{3^{502}}+\frac{2019}{5^{503}}\)
=> 3A-A=(\(7+\frac{11}{3}+\frac{15}{3^2}+\frac{19}{3^3}+...+\frac{2015}{3^{502}}+\frac{2019}{5^{503}}\))-(\(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+\frac{19}{3^4}+...+\frac{2015}{3^{503}}+\frac{2019}{3^{504}}\))
2A=\(7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+\left(\frac{19}{3^3}-\frac{15}{3^3}\right)+...+\left(\frac{2019}{3^{503}}-\frac{2015}{3^{503}}\right)-\frac{2019}{3^{504}}\)
2A=\(7+\frac{4}{3}+\frac{4}{3^2}+\frac{4}{3^3}+...+\frac{4}{3^{503}}-\frac{2019}{3^{504}}\)
=> A=\(\frac{7}{2}+2\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{503}}\right)-\frac{2019}{2.3^{504}}\)
Em làm tiếp Xét
B=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{503}}\)
3B=\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{502}}\)
=> 3B-B=\(1-\frac{1}{3^{503}}\)
=> B=\(\frac{1}{2}-\frac{1}{2.3^{503}}\)
=> A=\(\frac{7}{2}+2\left(\frac{1}{2}-\frac{1}{2.3^{503}}\right)-\frac{2019}{2.3^{504}}=\frac{9}{2}-\frac{1}{3^{503}}-\frac{2019}{2.3^{504}}< \frac{9}{2}\)
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)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
25 tháng 9 2020
a/
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(A=2A-A=1-\frac{1}{2^{100}}< 1\)
b/
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)
27 tháng 12 2018
\(A=2\cdot\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2017^2}\right)< 2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\right)\)
Đặt \(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(\Rightarrow M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(\Rightarrow M=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}< \frac{1}{1009}+\frac{1}{1009}+...+\frac{1}{1009}\)(1008 số hạng )
hay\(M< \frac{1008}{1009}\Rightarrow A< 2\cdot\frac{1008}{1009}=\frac{504}{1009}\left(ĐPCM\right)\)
Chứng tỏ A < \(\frac{9}{2}\)