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NH
0
DT
1
JK
26 tháng 7 2019
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}< \frac{1}{2\cdot3}\); \(\frac{1}{4^2}< \frac{1}{3\cdot4}\); ....; \(\frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow S< 1-\frac{1}{9}\)
\(\Rightarrow S< \frac{8}{9}\) (1)
\(\frac{1}{2^2}>\frac{1}{2\cdot3};\frac{1}{3^2}>\frac{1}{3\cdot4};\frac{1}{4^2}>\frac{1}{4\cdot5};...;\frac{1}{9^2}>\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow S>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow S>\frac{2}{5}\) (2)
(1)(2) => 2/5 < S < 8/9
26 tháng 7 2019
\(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}< \frac{1}{a^2}\)
\(\frac{1}{a}-1-\frac{1}{a}=-1< \frac{1}{a^2}\) Vì \(\frac{1}{a^2}>0;-1< 0\)
Khi đó thì ĐỀ SAI
KA
1
LP
7 tháng 5 2019
A=\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{98^2}\)
A=\(\frac{1}{3.3}\)+\(\frac{1}{4.4}\)+\(\frac{1}{5.5}\)+...+\(\frac{1}{98.98}\)
A<\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{97.98}\)=\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{97}\)-\(\frac{1}{98}\)=\(\frac{1}{2}\)-\(\frac{1}{98}\)=\(\frac{24}{49}\)<1.
Vậy A<1
NL
5
24 tháng 6 2020
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
Vậy \(A< 1\)
24 tháng 6 2020
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{100^2}=\frac{1}{100\cdot100}< \frac{1}{99\cdot100}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
=> \(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
Lại có : \(\frac{99}{100}< 1\)
=> \(A< \frac{99}{100}< 1\)=> \(A< 1\)( đpcm )
S
24 tháng 9 2016
A = 1+2+22+...+210
=> 2A = 2+22+23+...+211
=> 2A - A = (2+22+23+...+211) - (1+2+22+...+210)
=> A = 211 - 1
B = 1+3+32+...+310
=> 3B = 3+32+33+...+311
=> 3B - B = (3+32+33+...+311) - (1+3+32+...+310)
=> 2B = 311 - 1
=> B = \(\frac{3^{11}-1}{2}\)
AH
24 tháng 9 2016
A = 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10
2A = 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11
2A - A = ( 2 + 2 2 + 2 3 + 2 4 + ... + 2 10 + 2 11 )
- ( 1 + 2 1 + 2 2 + 2 3 + ... + 2 9 + 2 10 )
A = 2 11 - 1
A = 2047
B = 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10
3B = 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11
3B - B= ( 3 1 + 3 2 + 3 3 + 3 4 + ... + 3 10 + 3 11 )
- ( 1 + 3 1 + 3 2 + 3 3 + ... + 3 9 + 3 10 )
2B = 3 11 - 1
B = \(\frac{3^{11}-1}{2}\)
B = 88573
Ta có:
\(A=\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{1000}\right)^2< 1\)
\(A=\frac{1}{4}+\frac{1}{9}+...+\frac{1}{1000000}< 1\)
\(\frac{1}{4}< \frac{1}{1\cdot2}\)
\(\frac{1}{9}< \frac{1}{2\cdot3}\)
\(...\)
\(\frac{1}{1000000}< \frac{1}{999.1000}\)
\(A< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{999\cdot1000}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}\)
\(A< \frac{1}{1}-\frac{1}{1000}< 1\)
\(\Rightarrow A< 1\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{999.1000}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{999}-\frac{1}{1000}\)
\(A< 1-\frac{1}{1000}\)
\(=>A< 1\)
\(=>ĐPCM\)