Chứng tỏ rằng:
1)1-\(\frac{1}{2}\)-\(\frac{1}{2^2}\)-\(\frac{1}{2^3}\)-\(\frac{1}{2^4}\)-...-\(\frac{1}{2^{10}}\)>\(\frac{1}{2^{11}}\)
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Khách
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Những câu hỏi liên quan
21 tháng 9 2015
a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm
b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}
LH
11 tháng 8 2016
Ta có:
A = \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\)+....+ \(\frac{1}{11^2}\)
A = \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + \(\frac{1}{4.4}\)+....+ \(\frac{1}{11.11}\)
A < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) + .... + \(\frac{1}{10.11}\)
A < 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ...... + \(\frac{1}{10}\) - \(\frac{1}{11}\)
A < 1 - \(\frac{1}{11}\)
\(\Rightarrow\) A < \(\frac{10}{11}\)
11 tháng 8 2016
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.........
\(\frac{1}{11^2}< \frac{1}{10.11}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)
Lại có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{10}{11}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{11^2}< \frac{10}{11}\) ( đpcm )
KH
1
DL
9 tháng 5 2017
Bài này nhiều người đăng lắm,bạn vào câu hỏi tương tự
Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Đặt A =\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{3\cdot2}\)
...
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}< 1\)
\(\Rightarrow B< A< 1\left(đpcm\right)\)
NT
3
10 tháng 5 2017
Đặt A=đã cho.
Ta thấy:
1/2^2<1/1*2(vì 2^2>1*2).
1/3^2<1/2*3(vì 3^2>2*3).
...
1/10^2<1/9*10(vì 10^2>9*10).
=>A<1/1*2+1/2*3+1/3*4+...+1/9*10.
=>A<1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10.
=>A<1-1/10.
=>A<9/10.
Mà 9/10<1.
=>A<1.
Vậy A<1(đpcm).
VP
2
22 tháng 3 2018
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
....
\(\frac{1}{10^2}\)< \(\frac{1}{9.10}\)
=> \(\frac{1}{2^2}+....+\frac{1}{10^2}\)< \(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\)
=> \(\frac{1}{2^2}+....+\frac{1}{10^2}\)< \(\frac{9}{10}\)< 1
=> \(\frac{1}{2^2}+....+\frac{1}{10^2}\)< 1 ( dpcm )
HN
13 tháng 9 2015
1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)
19 tháng 4 2016
Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)
Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)
Vậy B < 1
HT
10 tháng 6 2016
Ta có \(D=\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}.\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}< 1\)
\(\Rightarrow D< 1\)
Vậy \(D< 1\)
10 tháng 6 2016
Ta có: 1/22 < 1/1.2
1/32 < 1/2.3
1/42 < 1/3.4
......
1/102 < 1/9.10
=> D < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10
=> D < 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 -1/10
=> D < 1 - 1/10
=> D < 9/10
=. D < 9/10 < 1
=> D < 1 ( đpcm )
\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)(1)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}\)
Thay A vào (1)
\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)
\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Ta có: 210 < 211
\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)