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4A = 1 +1/2^2+1/2^4+....+1/2^98
3A = 4A-A = (1+1/2^2+1/2^4+....+1/2^98) - (1/2^2+1/2^4+....+1/2^100) = 1 - 1/2^100 < 1
=> A < 1/3 ( ĐPCM )
k mk nha
a, \(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2C=1-\frac{1}{3^{99}}\)
\(C=\frac{1}{2}-\frac{1}{2.3^{99}}< \frac{1}{2}\)(đpcm)
b, Đặt \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6A=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6A-2A=\left(3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4A=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4A=3-\frac{397}{3^{100}}\)
\(A=\frac{3}{4}-\frac{397}{4.3^{100}}< \frac{3}{4}\)(đpcm)
Ta có: \(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}\)
\(\frac{2}{3}A=\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{100}{3^{101}}\right)\)
\(\frac{2}{3}A=\frac{1}{3}+\left(\frac{2}{3^2}-\frac{1}{3^2}\right)+\left(\frac{3}{3^3}-\frac{2}{3^3}\right)+...+\left(\frac{100}{3^{100}}-\frac{99}{3^{100}}\right)-\frac{100}{3^{101}}\)
\(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}-\frac{100}{3^{101}}\)
Đặt: \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
\(\Rightarrow\frac{1}{3}B=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\)
\(\Rightarrow\frac{2}{3}B=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{101}}\right)\)
\(=\frac{1}{3}-\frac{1}{3^{101}}\)\(\Leftrightarrow B=\left(\frac{1}{3}-\frac{1}{3^{101}}\right):\frac{2}{3}=\left(\frac{1}{3}-\frac{1}{3^{101}}\right).\frac{3}{2}\)
Thay \(B\) vào \(\frac{2}{3}A\), ta có: \(\frac{2}{3}A=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100}{3^{101}}\)
\(\Rightarrow A=\left[\frac{3}{2}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{100}{3^{101}}\right]:\frac{2}{3}=\frac{9}{4}\left(\frac{1}{3}-\frac{1}{3^{101}}\right)-\frac{150}{3^{101}}\)
\(A=\frac{3}{4}-\frac{9}{4}.\frac{1}{3^{101}}-\frac{150}{3^{101}}\Rightarrow A< \frac{3}{4}\)
Vậy \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)(ĐPCM)
Xong.
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
1>1/1*2
1/22>1/2*3
1/32>1/3/4
.....................
1/1002>1/100*101
=>1-1/22-...-1/1002>1/1*2-1/2*3-.....-1/100*101=1-1/2-1/2+1/3-1/3+......-1/100+1/101=1/101
vậy 1-1/22-....-1002
study well
k nha
ai k đúng cho mk thì mk trả lại gấp đôi và ngược lại
ai ghé qua nhớ để lại 1 k đúng
ủng hộ mk nha
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
M = \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}\)
M = 1 - (\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\))
Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)= \(1-\frac{1}{100}\)
M > 1 - (1 - \(\frac{1}{100}\)) =\(\frac{1}{100}\) (đpcm)
cảm ơn bn