Chứng minh rằng:
\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)
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AH
Akai Haruma
Giáo viên
25 tháng 1 2018
Lời giải:
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(A+3A=1+\frac{1-2}{3}+\frac{-2+3}{3^2}+\frac{3-4}{3^3}+\frac{-4+5}{3^4}+...+\frac{99-100}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-.....+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(4A=(1-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{3^3})+...+(\frac{1}{3^{98}}-\frac{1}{3^{99}})-\frac{100}{3^{100}}\)
\(4A=\frac{2}{3}+\frac{2}{3^3}+...+\frac{2}{3^{99}}-\frac{100}{3^{100}}\)
\(2A=\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{50}{3^{100}}\)
\(18A=3+\frac{1}{3}+...+\frac{1}{3^{97}}-\frac{450}{3^{100}}\)
\(\Rightarrow 18A-2A=3-\frac{1}{3^{99}}-\frac{450}{3^{100}}+\frac{50}{3^{100}}=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}\)
\(\Leftrightarrow 16A=3-\frac{1}{3^{99}}-\frac{400}{3^{100}}<3\Rightarrow A< \frac{3}{16}\)
VD
2 tháng 5 2018
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+...+99/3^98-100/3^99
3A+A=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99-100/3^100
<1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99
Đặt S=1-1/3+1/3^2-1/3^3+1/3^4-...+1/3^98-1/3^99
3S=3-1+1/3-1/3^2+1/3^3-...-1/3^98
3S+S=3-1/3^99
S=(3-1/3^99) :4
S=3/4-1/4.3^99
\(\Rightarrow\)4A<3/4-1/4.3^99
\(\Rightarrow\)A<(3/4-1/4.3^99):4
\(\Rightarrow\)A<3/16-1/16.3^99<3/16
Vậy 1/3-2/3^2+3/3^3-4/3^4+...+99/3^99-100/3^100<3/16
MV
22 tháng 8 2017
\(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\\ 3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\\ 3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\\ 2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)
\(6A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\\ 6A-2A=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\\ 4A=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\\ A=\dfrac{3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{4}=\dfrac{3}{4}-\dfrac{\dfrac{101}{3^{99}}}{4}-\dfrac{\dfrac{100}{3^{100}}}{4}< \dfrac{3}{4}\)
Vậy ...
Đặt A = \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
\(2A=3A-A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)Đặt B= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3B=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(2B=3B-B=3-\dfrac{1}{3^{99}}\)
Nhận xét : 2B < 3 => B < \(\dfrac{3}{2}\)
=> \(B-\dfrac{100}{3^{100}}< \dfrac{3}{2}\) hay 2A < \(\dfrac{3}{2}\)
=> Đpcm
***tik mik nhé***
Ta có :
3M=1+2/3+3/3^2+...+100/3^99
Suy ra :
2M=1+(1/3+1/3^2+1/3^3+...+1/3^99)-100/3^100
Xét B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)
3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)
2B=1-\(\dfrac{1}{3^{99}}\)<1/2
Suy ra : 2M<1+1/2 nên M<3/4