cho A= 1/2^2 + 1/3^2 + ... + 1/2017^2. Chứng minh A<3/4
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H
1
29 tháng 3 2017
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}+\frac{1}{2017^2}\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2016.2016}+\frac{1}{2017.2017}\)
Ta thấy \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};\frac{1}{4.4}< \frac{1}{3.4};...;\frac{1}{2016.2016}< \frac{1}{2016.2017};\frac{1}{2017.2017}< \frac{1}{2017.2018}\)
Suy ra \(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}+\frac{1}{2017.2018}\)
Nên \(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-...+\frac{1}{2017}-\frac{1}{2018}\)
Khi đó \(A< 1-\frac{1}{2018}< 1\)nên A < 1
Suy ra A - 1 < 0
Vậy A - 1 < 0
MN
1
MV
5 tháng 5 2017
\(A=\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\\ 2A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\\ 2A-A=\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2^2}+\dfrac{2}{2^3}+\dfrac{3}{2^4}+...+\dfrac{2016}{2^{2017}}\right)\\ A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ 2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\\ 2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}-\dfrac{2016}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}-\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{2017}{2^{2016}}-\dfrac{2016}{2^{2017}}\\ A=1-\dfrac{4034}{2^{2017}}-\dfrac{2016}{2^{2017}}\\ A=1-\left(\dfrac{4034}{2^{2017}}+\dfrac{2016}{2^{2017}}\right)\\ A=1-\dfrac{6050}{2^{2017}}< 1\)
Vậy \(A< 1\)
HV
0
N
1
KN
16 tháng 4 2019
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A< 1-\frac{1}{2017}=\frac{2016}{2017}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}< \frac{2016}{2017}\left(đpcm\right)\)
HV
0
DN
0
LH
1
KT
1 tháng 12 2018
\(N=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)
=> \(3N=1+\frac{1}{3}+...+\frac{1}{3^{2017}}\)
=> \(3N-N=\left(1+\frac{1}{3}+...+\frac{1}{3^{2017}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\right)\)
<=> \(2N=1-\frac{1}{3^{2018}}< 1\)
<=> \(N< \frac{1}{2}\)
=> dpcm
S
14 tháng 5 2017
a, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2017^2}< \frac{1}{2016.2017}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2017^2}>\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}=1-\frac{1}{2017}< 1\)Vậy...
b, Đặt A = \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{10000}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)
Thay B vào A ta được:
\(A< \frac{1}{4}\left(1+1\right)=\frac{1}{4}.2=\frac{1}{2}\)
Vậy....
S
14 tháng 5 2017
c, Ta có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};....;\frac{1}{9^2}>\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(1)
Lại có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{9^2}< \frac{1}{8.9}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) suy ra \(\frac{2}{5}< A< \frac{8}{9}\)(đpcm)
d, chắc là đề sai
e, giống câu a
PL
2
20 tháng 9 2018
ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2015^2}< \frac{1}{2014.2015};\frac{1}{2016^2}< \frac{1}{2015.1026};\frac{1}{2017^2}< \frac{1}{2016.2017}\)
=> 1/22 + 1/32 + 1/42 + ... + 1/20152 + 1/20162 + 1/20172 < 1/22 + (1/2.3 + 1/3.4 + ....+1/2014.2015 + 1/2015.2016 + 1/2016.2017)
= 1/4 + 1/2 - 1/2017 = 3/4- 1/2017 < 3/4
=> đ p c m
20 tháng 9 2018
ta có: \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{2015^2}< \frac{1}{2014.2015};\frac{1}{2016^2}< \frac{1}{2015.1026};\frac{1}{2017^2}< \frac{1}{2016.2017}\)
=> 1/22 + 1/32 + 1/42 + ... + 1/20152 + 1/20162 + 1/20172 < 1/22 + (1/2.3 + 1/3.4 + ....+1/2014.2015 + 1/2015.2016 + 1/2016.2017)
= 1/4 + 1/2 - 1/2017 = 3/4- 1/2017 < 3/4
=> đ p c m
Ta có : \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
.....................
\(\frac{1}{2017^2}< \frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=\frac{1}{4}+\frac{1}{2}-\frac{1}{2017}\)
\(A=\frac{3}{4}-\frac{1}{2017}\left(đpcm\right)\) . Vậy A < \(\frac{3}{4}\)