Chứng minh:
\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{2014}{3^{2014}}< \frac{1}{5}\)
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PN
8 tháng 5 2016
Ta có : A = 1/3 - 2/3^2 + 3/3^3 - 4/3^4 +...- 2014/3^2014
=> 3A = 1 - 2/3 + 3/3^2 - 4/3^3 +...- 2014/3^2013
=> 4A = 1- 1/3 + 1/3^2 -...- 1/3^2013 - 2014/3^2014
Xét B = 1-1/3+1/3^2 -...- 1/3^2013
=> 3B = 3 - 1 + 1/3 -...- 1/3^2012
=> 4B = 3- 1/3^2013
=> B = (3- 1/3^2013)/4 < 3/4
=> 4A < 3/4 - 2014/3^2014< 3/4
=> A < 3/16 < 3/15 =1/5
Vậy A < 1/5 (đpcm)
Chúc bạn học tốt
MN
1
20 tháng 6 2019
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{2014}{2^{2014}}\)
\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2014}{2^{2013}}\)
\(\Rightarrow2A-A=\left(1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{2014}{2^{2013}}\right)-\left(\frac{1}{2}+\frac{2}{2^2}+...+\frac{2014}{2^{2014}}\right)\)
\(\Rightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}-\frac{2014}{2^{2014}}\)
Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
\(\Rightarrow2B=2+1+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2B-B=\left(2+1+...+\frac{1}{2^{2012}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2013}}\right)\)
\(\Rightarrow B=2-\frac{1}{2^{2013}}< 2\)
\(\Rightarrow B< 2\)
\(\Rightarrow A< 2-\frac{2014}{2^{2014}}< 2\)
\(\Rightarrow A< 2\left(đpcm\right)\)
CC
9 tháng 12 2019
Ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
Tương tự : \(\frac{1}{3^2}< \frac{1}{2.3}\); \(\frac{1}{4^2}< \frac{1}{3.4}\); ......... ; \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{2013.2014}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}=\frac{2013}{2014}\)
\(\Rightarrow S< \frac{2013}{2014}\left(đpcm\right)\)
Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...-\frac{2014}{3^{2014}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-....-\frac{2014}{3^{2013}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-...-\frac{2014}{3^{2013}}\right)+\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-....-\frac{2014}{3^{2014}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}-\frac{2014}{3^{2014}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}\)
3B = \(3-1+\frac{1}{3}-...-\frac{1}{3^{2012}}\)
3B + B = \(\left(3-1+\frac{1}{3}-...-\frac{1}{3^{2012}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{2013}}\right)\)
4B = \(3-\frac{1}{3^{2013}}\)
=> 4B < 3 => B < \(\frac{3}{4}\)(2)
Từ (1)(2) => 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)<\(\frac{1}{5}\)(dpcm)
Nhanh nha