1/2+1/2^2+1/2^3+....+1/2^2013<1
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BC
0
DD
4
3 tháng 1 2018
Ta có :
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(=1-\frac{1}{2013}< 1\)( đpcm )
S
3 tháng 1 2018
\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
....
\(\frac{1}{2013^2}< \frac{1}{2012.2013}=\frac{1}{2012}-\frac{1}{2013}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=1-\frac{1}{2013}< 1\)
NN
6 tháng 1 2018
Gọi biểu thức trên là A.
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}.\)
Ta thấy:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}.\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}.\)
..................
\(\dfrac{1}{2013^2}=\dfrac{1}{2012.2013}.\)
\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}\)
\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}.\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}.\)
\(=1-\dfrac{1}{2013}.\)
\(< 1\left(đpcm\right).\)
Vậy \(A< 1.\)
LT
1
3 tháng 1 2018
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
....................
\(\dfrac{1}{2013^2}< \dfrac{1}{2012.2013}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{2012.2013}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{2013^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{2013^2}< 1-\dfrac{1}{2013}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{2013^2}< 1\left(đpcm\right)\)
LT
2
NT
3 tháng 1 2018
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{2013^2}\\ =\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{2013.2013}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}\\ -1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}\\ =1-\dfrac{1}{2013}< 1\)
3 tháng 1 2018
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}=1-\dfrac{1}{2}+\dfrac{1}{2}+....+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(\Rightarrow dpcm\)
LT
1
3 tháng 1 2018
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2012.2013}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}=1-\dfrac{1}{2013}< 1\left(đpcm\right)\)
AT
0
RP
1
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
Khi đó \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\right)\)
=> \(A=1-\frac{1}{2^{2013}}< 1\left(\text{Đpcm}\right)\)