CMR:A=1/1^2+1/2^2+...+1/100^2 <1 3/4
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ND
2
DP
19 tháng 7 2017
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{101}\)
\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{99}{202}< \frac{3}{4}\)
\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)
19 tháng 7 2017
A= 1/4 +1/3^2 +1/4^2 +.....+ 1/100^2
< 1/4 + 1/2.3 + 1/3.4 +.....+1/99.100
=1/4 + 1/2-1/3+1/3-1/4+......+1/99-1/100
=1/4 +1/2 - 1/100 < 1/4+1/2 = 3/4
=> ĐPCM
NT
4
25 tháng 5 2016
A= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)
A <\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
A<\(1-\frac{1}{n}\)=\(\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}< 1\)
Vậy A < 1
25 tháng 5 2016
Ta có:
1/22 < 1/1.2
1/32 < 1/2.3
1/42 < 1/3.4
..................
=> 1/n2 < 1/n(n-1)
=> 1/22 + 1/32 + 1/42 + ... + 1/n2 < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/n(n-1)
=> A < 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/n-1 + 1/n
=> A < 1 - 1/n
Vơi n thuộc N* => 1 - 1/n < 1 ( vì 1/n lúc đó lớn hơn 0 )
=> A < 1 - 1/n < 1
đpcm
CC
6 tháng 3 2020
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+........+\frac{100}{2^{100}}\)
\(\Rightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+..........+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}+\frac{100}{2^{100}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{98}}+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=A=2-\frac{100}{2^{99}}< 2\)
Vậy \(A< 2\)
NT
2
24 tháng 4 2019
ta có:1/2^2=1/4
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
...
1/100^2<1/99.100=1/99-1/100
=> A=1/2^2+1/3^2+1/4^2+.....+1/100^2<1/4+1/2-1/3+1/3-1/4+...+1/99-1/100
<1/4+1/2-1/100<1/2
24 tháng 4 2019
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{100^2}\)
\(< \frac{1}{4}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}\)
\(< \frac{1}{2}-\frac{1}{100}\)
\(< \frac{1}{2}\)
SY
19 tháng 3 2016
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.