cho: D= 1/2^2 + 1/3^2 + ... + 1/100^2
CMR: D<3/4
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LT
0
5 tháng 2 2022
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
NM
1
NH
20 tháng 8 2020
\(D=\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{99^2}+\frac{1}{100^2}< \frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)(1)
\(D=\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}>\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{4}-\frac{1}{101}>\frac{1}{5}\)(2)
Từ (1) và (2) :
\(\Rightarrow\frac{1}{5}< D< \frac{1}{3}\)( đpcm )
NP
1
S
27 tháng 6 2018
\(D=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+....+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(\Rightarrow2D=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+....+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(\Rightarrow2D+D=1-\frac{1}{2^{100}}\)
\(\Rightarrow3D=1-\frac{1}{2^{100}}\)
\(\Rightarrow D=\frac{1-\frac{1}{2^{100}}}{3}\)
\(\Rightarrow D=\frac{1}{3}-\frac{1}{3.2^{100}}\)
Chúc bạn hok tốt!
NT
1
22 tháng 4 2017
Ta có :
\(D=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+..............+\dfrac{100}{3^{100}}+\dfrac{101}{3^{101}}\)
\(3D=1+\dfrac{2}{3}+\dfrac{3}{3^2}+.............+\dfrac{100}{3^{99}}\)
\(3D-D=\left(1+\dfrac{2}{3}+\dfrac{3}{3^3}+.....+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+.......+\dfrac{101}{3^{101}}\right)\)
\(2D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+............+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6D=3+1+\dfrac{1}{3}+............+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)
\(6D-2D=\left(3+1+\dfrac{1}{3}+..........+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+......+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)\(4D=3-\dfrac{100}{3^{99}}-\dfrac{1}{3^{99}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{300}{3^{100}}-\dfrac{3}{3^{100}}+\dfrac{100}{3^{100}}\)
\(4D=3-\dfrac{203}{3^{100}}< 3\)
\(\Rightarrow D< \dfrac{3}{4}\rightarrowđpcm\)
~ Học tốt ~
MN
0
Ta có D = 1/2^2 + 1/3^2 + ... + 1/100^2
= 1/4 + ( 1/3.3 + ... + 1/100.100 )
Ta thấy
1/3.3 < 1/2.3
...
1/100.100 < 1/99.100
Suy ra 1/4 + 1/3.3 + ... + 1/100.100 < 1/4 + 1/2.3 + ... + 1/99.100
Suy ra 1/4 + 1/3^2 + ... + 1/100^2 < 1/4 + ( 1/2 - 1/3 ) + ... + ( 1/99 - 1/100 )
Hay D < 1/4 + ( 1/2 - 1/100 ) + { ( 1/3 + ... +1/99 ) - (1/3 + ... + 1/99 ) }
Suy ra D < 1/4 + 1/2 -1/100 + 0
Suy ra D < 3/4 - 1/100
Do đó D < 3/4