cho biểu thức: A=1+1/2 +1/2^2 +1/2^3 +.......+1/2^2017
Chứng tỏ A<1
K
Khách
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DT
2
6 tháng 5 2018
\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
\(\Rightarrow\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}< 1\left(đpcm\right)\)
6 tháng 5 2018
ta có: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)
\(A=1-\frac{1}{2^{100}}< 1\)
\(\Rightarrow A< 1\left(đpcm\right)\)
DT
1
TH
2
1 tháng 5 2018
A= \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+...+ \(\frac{1}{2^{99}}\)+ \(\frac{1}{2^{100}}\).
2A= 1+ \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+...+ \(\frac{1}{2^{100}}\)+ \(\frac{1}{2^{101}}\).
2A- A=( 1+ \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+...+ \(\frac{1}{2^{100}}\)+ \(\frac{1}{2^{101}}\))-( \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+...+ \(\frac{1}{2^{99}}\)+ \(\frac{1}{2^{100}}\)).
A= 1- \(\frac{1}{2^{100}}\)< 1.
=> A< 1.
Vậy A< 1.
WH
1 tháng 5 2018
Ta có
\(2A=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow2A=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+...+\frac{2}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
\(\Rightarrow A< 1\)
Vậy A<1 (đpcm)
NT
1
9 tháng 5 2018
Ta có:3.A=1+1/3+1/3^2+...+1/3^97 +1/3^98
=>3.A - A=(1+1/3+1/3^2+...+1/3^98 + 1/3^99)-(1/3+1/3^2 +1/3^3+...+1/3^98+1/3^99)
=>2.A=1-1/3^99
=>A=1/2 -1/3^99.1/2 <1/2
Vậy ... T I C K cho mình với nha
1 tháng 1
Câu 1:
b: ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
\(\dfrac{1}{x-3}-\dfrac{1}{x+3}+\dfrac{2x}{9-x^2}\)
\(=\dfrac{1}{x-3}-\dfrac{1}{x+3}-\dfrac{2x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x+3-x+3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=-\dfrac{2}{x+3}\)
c: ĐKXĐ: \(x\notin\left\{2;0\right\}\)
Sửa đề: \(\dfrac{x+1}{x-2}+\dfrac{4-5x}{x^3+4x}:\dfrac{x-2}{x^2+4}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x^2+4\right)}\cdot\dfrac{x^2+4}{x-2}\)
\(=\dfrac{x+1}{x-2}+\dfrac{4-5x}{x\left(x-2\right)}\)
\(=\dfrac{x\left(x+1\right)+4-5x}{x\left(x-2\right)}=\dfrac{x^2+x-5x+4}{x\left(x-2\right)}\)
\(=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
MH
1
22 tháng 4 2016
Bạn xem lời giải của mình nhé:
Giải:
A luôn > 0 (vì các số hạng trong tổng A đều lớn hơn 0)(1)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\\ 2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\\ 2A-A=1-\frac{1}{2^{100}}< 1\)
\(A< 1\)(2)
Từ (1) và (2) \(\Rightarrow0< A< 1\left(đpcm\right)\)
Chúc bạn học tốt!
Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(2A=2+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2017}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{2}{2^{2016}}\right)\)
\(A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2016}}-\frac{1}{2^{2017}}\)
\(A=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Vậy A < 1
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(A=2-\frac{1}{2^{2017}}\left(đpcm\right)\)