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\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)
\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)
\(8A=1-3=-2\)
A=\(\frac{-2}{8}=\frac{-1}{4}\)
\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)
Vậy B=1
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
\(A=\dfrac{-1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{-1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)
Cộng vế với vế:
\(A+\dfrac{1}{3}A=\dfrac{-1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)
\(\Rightarrow\dfrac{4}{3}A=\dfrac{-1}{3}+\dfrac{1}{3^{101}}\)
\(\Rightarrow A=\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)\)
Do \(\dfrac{1}{3^{100}}< \dfrac{1}{3}< 1\Rightarrow A< 0\)
\(\Rightarrow\left|A\right|=-A=-\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)=\dfrac{1}{4}\left(1-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow B=4\left|A\right|+\dfrac{1}{3^{100}}=1-\dfrac{1}{3^{100}}+\dfrac{1}{3^{100}}=1\)
a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)