Cho A=1/2²+1/3²+1/4²+...+1/2013²
Chứng minh: A<3/4
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17 tháng 4 2017
Ta có :
\(A=\dfrac{1}{1+3}+\dfrac{1}{1+3+5}+...........+\dfrac{1}{1+3+.....+2013}\)
\(A=\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}+\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}+.........+\dfrac{1}{\dfrac{\left(1+2013\right).1007}{2}}\)
\(A=\dfrac{2}{2.4}+\dfrac{2}{3.6}+\dfrac{2}{4.8}+...........+\dfrac{2}{1007.2014}\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+..........+\dfrac{1}{1007.1007}\)
\(\Rightarrow A< \dfrac{1}{2.2}+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{1006.1008}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...........+\dfrac{1}{1006}-\dfrac{1}{1007}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\left(\dfrac{1}{2}-\dfrac{1}{1007}\right)\)
\(\Rightarrow A< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\) \(\rightarrowđpcm\)
~ Chúc bn học tốt ~
BC
0
P
0
26 tháng 4 2019
a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< 1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A< 1+\left(1-\frac{1}{100}\right)\Rightarrow A< 1+1-\frac{1}{100}\Rightarrow A< 2-\frac{1}{100}\Rightarrow A< 2\left(ĐPCM\right)\)
b, \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(\Rightarrow B< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2011\cdot2012}\)
\(\Rightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(\Rightarrow B< 1-\frac{1}{2012}\Rightarrow B< 1\left(1\right)\)
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2012^2}\)
\(\Rightarrow B>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2012\cdot2013}\)
\(\Rightarrow B>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(\Rightarrow B>\frac{1}{2}-\frac{1}{2013}\Rightarrow\frac{1}{2}-\frac{1}{2013}< B\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{2}-\frac{1}{2013}< B< 1\)
DT
26 tháng 4 2019
a)A=1+1/22+1/32+....+1/1002
<1+1/1.2+1/2.3+...+1/99.100=1+1-1/2+1/2-1/3+...+1/99-1/100=2-1/100=199/200<2
b)B=1/22+1/32+...+1/20122
<1/1.2+1/2.3+...+1/2011.2012=1-1/2+1/2-1/3+...+1/2011-1/2012=1-1/2012=2011/2012
1/2-1/2013=2011/4026<2011/2012<1
23 tháng 5 2022
\(A=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2+...+\left(\dfrac{1}{2013}\right)^2\)
\(A=\left(\dfrac{1}{2+3+4+...+2013}\right)^2\)
\(A=\left(\dfrac{1}{\left(2013-2\right)+1}\right)^2\)
\(A=\left(\dfrac{1}{2012}\right)^2\)
\(A=\dfrac{1}{2012\cdot2012}\)
\(\Rightarrow A=\dfrac{1}{2012}< \dfrac{3}{4}\)
PT
0
A< \(\frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2012+2013}\)
⇔ A< \(\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2012}-\frac{1}{2013}\)
⇔ A<\(\frac{1}{4}+\frac{1}{2}-\frac{1}{2013}\)
vì \(\frac{1}{4}+\frac{1}{2}-\frac{1}{2013}< \frac{3}{4}\)
nên A < \(\frac{3}{4}\)