A=1/3+1/3^2+1/3^3+...+1/3^100
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14 tháng 11 2023
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
5 tháng 5 2015
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\RightarrowĐPCM\)
16 tháng 4 2017
a)Ta có :
\(A=\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+............+\dfrac{1}{4^{100}}\)
\(4A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+..........+\dfrac{1}{4^{99}}\)
\(4A-A=\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{99}}\right)-\left(\dfrac{1}{4}+\dfrac{1}{4^2}+.....+\dfrac{1}{4^{100}}\right)\)
\(3A=1-\dfrac{1}{4^{100}}\)
\(\Rightarrow A=\dfrac{1-\dfrac{1}{4^{100}}}{3}\)
~ Chúc bn học tốt ~
24 tháng 11 2021
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5 tháng 2 2022
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
PT
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