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4 tháng 8 2016
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+1-\frac{1}{100}\)
\(=2-\frac{1}{100}< 2\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
NM
0
EG
2
XO
8 tháng 2 2020
Ta có : A = \(\frac{1}{100^2}+\frac{1}{101^2}+...+\frac{1}{199^2}=\frac{1}{100.100}+\frac{1}{101.101}+...+\frac{1}{199.199}\)
> \(\frac{1}{100.101}+\frac{1}{101.102}+...+\frac{1}{199.200}=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{199}-\frac{1}{200}\)
= \(\frac{1}{100}-\frac{1}{200}=\frac{1}{200}\Rightarrow A>\frac{1}{200}\left(1\right)\)
Lại có : A = \(\frac{1}{100^2}+\frac{1}{101^2}+...+\frac{1}{199^2}=\frac{1}{100.100}+\frac{1}{101.101}+...+\frac{1}{199.199}\)
\(< \frac{1}{99.100}+\frac{1}{100.101}+...+\frac{1}{198.199}=\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+...+\frac{1}{198}-\frac{1}{199}\)
\(=\frac{1}{99}-\frac{1}{199}\Rightarrow A< \frac{1}{99}\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{200}< A< \frac{1}{99}\left(\text{ĐPCM}\right)\)
8 tháng 2 2020
Cho A=\(\frac{1}{100^2}+\frac{1}{101^2}+......................+\frac{1}{198^2}+\frac{1}{199^2}\)
CMR:\(\frac{1}{200}< A< \frac{1}{99}\)
+)Ta có:A=\(\frac{1}{100^2}+\frac{1}{101^2}+......................+\frac{1}{198^2}+\frac{1}{199^2}\)
=>A=\(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)
+)Ta thấy :\(\frac{1}{100.100}\)>\(\frac{1}{100.101}\)
\(\frac{1}{101.101}>\frac{1}{101.102}\)
.............................................
\(\frac{1}{198.198}>\frac{1}{198.199}\)
\(\frac{1}{199.199}>\frac{1}{199.200}\)
=> \(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)>\(\frac{1}{100.101}+\frac{1}{101.102}+................+\frac{1}{198.199}+\frac{1}{199.200}\)
=>A>\(\frac{1}{100.101}+\frac{1}{101.102}+................+\frac{1}{198.199}+\frac{1}{199.200}\)
=>A>\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+........+\frac{1}{198}-\frac{1}{199}+\frac{1}{199}-\frac{1}{200}\)
=>A>\(\frac{1}{100}-\frac{1}{200}=\frac{2}{200}-\frac{1}{200}=\frac{1}{200}\)
=>A>\(\frac{1}{200}\)(1)
+)Ta lại có:
A=\(\frac{1}{100^2}+\frac{1}{101^2}+......................+\frac{1}{198^2}+\frac{1}{199^2}\)
=>A=\(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)
+)Ta lại thấy:\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\frac{1}{101.101}< \frac{1}{100.101}\)
................................................
\(\frac{1}{198.198}< \frac{1}{197.198}\)
\(\frac{1}{199.199}< \frac{1}{198.199}\)
=>\(\frac{1}{100.100}+\frac{1}{101.101}+...........+\frac{1}{198.198}+\frac{1}{199.199}\)<\(\frac{1}{99.100}+\frac{1}{100.101}+.............+\frac{1}{197.198}+\frac{1}{198.199}\)
=>A<\(\frac{1}{99.100}+\frac{1}{100.101}+.............+\frac{1}{197.198}+\frac{1}{198.199}\)
=>A<\(\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}+...........+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\)
=>A<\(\frac{1}{99}-\frac{1}{199}\)
Mà A<\(\frac{1}{99}-\frac{1}{199}\)
=>A<\(\frac{1}{99}\)(2)
+)Từ (1) và (2)
=>\(\frac{1}{200}< A< \frac{1}{99}\)(ĐPCM)
Vậy \(\frac{1}{200}< A< \frac{1}{99}\)
Chúc bn học tốt
DN
1
DC
19 tháng 12 2019
B=1/2+(1/2)^2+................+(1/2)^100
=>1/2B=(1/2)^2+(1/2)^3+............+(1/2)^101
=>1/2B-B=(1/2^2+..............+1/2^101)-(1/2+..............+1/2^100)
=>1/2B-B=1/2^2+..............+1/2^101-1/2-..............-1/2^100
=>1/2B-B=1/2^101+(1/2^2-1/2^2)+................+(1/2^100-1/2^100)-1/2
=>1/2B-B=1/2^101+0+............+0-1/2
=>-1/2B=1/2^101-1/2
=>B=1/2^101-1/2
__________
-1/2
=>B<1
KM
1
KL
10 tháng 2 2019
Ta có A = 1/2+2/22+3/23+4/24+...+100/2100
<=> A = 1/2+2/4+3/9+4/16+...+100/2100
10 tháng 9 2018
2
a) (2x - 1)4 = 81
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)(1)
\(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{99}}\)(2)
lay (2)-(1)
\(2.A-A=A=1-\frac{1}{2^{100}}< 1\Rightarrow dpcm\)