Cho E=\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}.\)Chứng minh:C<\(\dfrac{5}{3}\)
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14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2023
Lời giải:
$M=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+....+\frac{100}{3^{100}}$
$3M=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}$
$\Rightarrow 2M=3M-M=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}$
$2M+\frac{100}{3^{100}}=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$
$3(2M+\frac{100}{3^{100}})=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}
$\Rightarrow 2(2M+\frac{100}{3^{100}})=3(2M+\frac{100}{3^{100}})-(2M+\frac{100}{3^{100}})=2-\frac{1}{3^{99}}$
$M=\frac{1}{2}-\frac{1}{4.3^{99}}-\frac{50}{3^{100}}<\frac{1}{2}< \frac{3}{4}$
Ta có đpcm.
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2023
Đặt \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)
\(\Rightarrow A+3A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow4A=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\) (1)
\(\Rightarrow12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\) (2)
Cộng vế (1) và (2):
\(\Rightarrow16A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\)
\(\Rightarrow16A< 3\)
\(\Rightarrow A< \dfrac{3}{16}\)
2 tháng 3 2023
Đặt `A` `=` `1/3 - 2/3^2+3/3^3 - 4/3^4+ ... + 99/3^99-100/3^100`
`=>3A=1 -2/3 +3/3^2 - 4/3^3+ ... - 100/3^99`
`=>4A=A+3A=1-1/3+1/3^2-1/3^3+...-1/3^99 - 100/3^100`
`=>12A=3.4A=3-1+1/3-1/3^2+...-1/3^98 - 100/3^99`
`=>16A=12A+4A=3-1/3^99-100/3^99-100/3^1...`
`=>16A=3-101/3^99-100/3^100`
`<=>A=3/16-(101/3^99+100/3^100)/16 < 3/16`
`=> A<3/16`
@Nae
C=13+232+333+434+...+1003100�=13+232+333+434+...+1003100
3C=1+23+332+433+...+1003993�=1+23+332+433+...+100399
3C−C=(1+23+332+433+...+100399)−(13+232+333+434+...+1003100)3�−�=(1+23+332+433+...+100399)−(13+232+333+434+...+1003100)
2C=1+13+132+133+...+1399−10031002�=1+13+132+133+...+1399−1003100
6C=3+1+13+132+...+1398−1003996�=3+1+13+132+...+1398−100399
6C−2C=(3+1+13+132+...+1398−100399)−(1+13+132+133+...+1399−1003100)6�−2�=(3+1+13+132+...+1398−100399)−(1+13+132+133+...+1399−1003100)
4C=3−100399−1399+10031004�=3−100399−1399+1003100
4C=3−3003100−33100+10031004�=3−3003100−33100+1003100
4C=3−2033100<34�=3−2033100<3
⇒C<34(đpcm)
k nha