Chứng minh rằng:
C = 1/3 + 2/32 + 3/33 + 4/34 + . . . . . . . + 100/3100 < 3/4
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S
1
S
0
TD
0
AD
0
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
VG
1
NM
1
NT
20 tháng 4 2018
\(A=3+3^2+3^3+3^4+.......+3^{100}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(\Rightarrow A=3.\left(1+3+3^2+3^3\right)+........+3^{97}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=3.40+.........+3^{97}.40\)
\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)
\(\Rightarrow A⋮40\)( 1 )
Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)
\(\Rightarrow A⋮120\)
Vậy \(A⋮120\)( ĐPCM )
MY
2
DN
16
28 tháng 7 2015
S có 30 số hạng. Nhóm thành 3 nhóm, mỗi nhóm 10 số hạng
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{42}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(S<\left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)
\(S<\frac{10}{30}+\frac{10}{40}+\frac{10}{50}\) ; \(S<\frac{47}{60}<\frac{48}{60}=\frac{4}{5}\) (1)
\(S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(S<\frac{10}{40}+\frac{10}{50}+\frac{10}{60}\) ; \(S>\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\) (2)
Từ (1) và (2) => \(\frac{3}{5}\)<S<\(\frac{4}{5}\)
\(C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3C-C=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6C=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6C-2C=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4C=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4C=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4C=3-\frac{203}{3^{100}}< 3\)
\(\Rightarrow C< \frac{3}{4}\left(đpcm\right)\)