A= 1/3 +2/3^2 + 3/3^3 + 4/3^4 +...+ 2012/3^2012
Chứng minh rằng : A <3/4
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HH
0
NT
0
L
5
3 tháng 2 2022
\(A=\left(3+3^2+3^3+3^4\right)+3^4\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(=120+3^4.120+...+3^{2008}.120=120\left(1+3^4+...+3^{2008}\right)⋮120\)
3 tháng 2 2022
\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(A=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
\(A=\left(3+3^2+3^3+3^4\right)\left(1+3^4+...+3^{2008}\right)\)
\(A=120\left(1+3^4+...+3^{2008}\right)⋮120\)
NP
0
PN
1
7 tháng 1 2021
Ta có: \(A=3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=3.40+3^5.40+...+3^{2009}.40\)
\(=120+3^4.120+...+3^{2008}.120\)
\(=120\left(1+3^4+...+3^{2008}\right)\)
Vì \(120⋮120\) nên \(120\left(1+3^4+...+3^{2008}\right)⋮120\)
hay \(A⋮120\) (đpcm)
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2012}{3^{2012}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{2012}{3^{2011}}\)
\(\Rightarrow3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{2012}{3^{2011}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{2012}{3^{2012}}\right)\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2011}}-\frac{2012}{3^{2012}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2010}}-\frac{2012}{3^{2011}}\)
\(\Rightarrow6A-2A=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2010}}-\frac{2012}{3^{2011}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2011}}-\frac{2012}{3^{2012}}\right)\)
\(\Rightarrow4A=3-\frac{2012}{3^{2011}}\)
\(\Rightarrow A=\frac{3-\frac{2012}{3^{2011}}}{4}=\frac{3}{4}-\frac{\frac{2012}{3^{2011}}}{4}=\frac{3}{4}-\frac{2012}{3^{2011}.4}\)
\(\Rightarrow A< \frac{3}{4}\)
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