\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
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10 tháng 4 2019
\(\)Đặt \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+...+\frac{1}{204}}=\frac{B}{C}\)
Biến đổi C:
\(C=\left(\frac{204}{1}+1\right)+\left(\frac{203}{2}+1\right)+\left(\frac{202}{3}+1\right)+...+\left(\frac{1}{204}+1\right)-204\)
\(=205+\frac{205}{2}+\frac{205}{3}+..+\frac{205}{204}+\frac{205}{205}-205\)
\(=205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}}{205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)}=\frac{1}{205}\)
SY
19 tháng 3 2016
a)Đặt A= \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{8}\) - \(\frac{1}{16}\) + \(\frac{1}{32}\) - \(\frac{1}{64}\) => A=\(\frac{1}{2^1}\) - \(\frac{1}{2^2}\) + \(\frac{1}{2^3}\) - \(\frac{1}{2^4}\) + \(\frac{1}{2^5}\) - \(\frac{1}{2^6}\)
=> 2A= 1-\(\frac{1}{2^1}\) + \(\frac{1}{2^2}\) - \(\frac{1}{2^3}\) + \(\frac{1}{2^4}\) - \(\frac{1}{2^5}\)
=> 3A= 1- \(\frac{1}{2^6}\) <1 => A<\(\frac{1}{3}\) => đpcm.
b) Đặt B=\(\frac{1}{3}\) - \(\frac{2}{3^2}\) + \(\frac{3}{3^3}\) - \(\frac{4}{3^4}\) +..+ \(\frac{99}{3^{99}}\) - \(\frac{100}{3^{100}}\)
=> 3B=1-\(\frac{2}{3}\) + \(\frac{3}{3^2}\) - \(\frac{4}{3^3}\) +...+\(\frac{99}{3^{98}}\) - \(\frac{100}{3^{99}}\)
=> 4B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) - \(\frac{100}{3^{99}}\) < 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\) (1)
Đặt B= 1-\(\frac{1}{3}\) + \(\frac{1}{3^2}\) - \(\frac{1}{3^3}\) +...+\(\frac{1}{3^{99}}\)
=> 3B= 3-1+\(\frac{1}{3}\) - \(\frac{1}{3^2}\) + \(\frac{1}{3^3}\) - \(\frac{1}{3^4}\) +...+ \(\frac{1}{3^{98}}\)
=> 4B= 3-\(\frac{1}{3^{99}}\) <3 => B<\(\frac{3}{4}\) (2)
=> 4A<B<\(\frac{3}{4}\) => A<\(\frac{3}{16}\) => đpcm.
17 tháng 6 2015
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
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17 tháng 6 2015
A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(9A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)
\(9A+A=\left(\frac{1}{3}-\frac{1}{3^{^2}}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(10A=\frac{1}{3}-\frac{1}{3^{204}}\)
A = (1/3 - 1/3204) : 10
Vậy A = (1/3 - 1/3204) : 10.
A= \(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\left(1\right)\\ \)
\(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\left(2\right)\)
Từ (1) và (2) vế theo vế ta có :\(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)
\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{206}}\right):\frac{8}{9}\)