Question

[sunnie]

cho biểu thức: \(A=\left(\frac{-1}{3}\right)+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}\)

tính \(B=4\left|A\right|+\frac{1}{3^{100}}\)

Answer 1

[Nguyễn Việt Lâm]

\(A=\dfrac{-1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow\dfrac{1}{3}A=\dfrac{-1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...-\dfrac{1}{3^{100}}+\dfrac{1}{3^{101}}\)

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 \(\dfrac{1}{3^{100}}< \dfrac{1}{3}< 1\Rightarrow A< 0\)

\(\Rightarrow\left|A\right|=-A=-\dfrac{1}{4}\left(\dfrac{1}{3^{100}}-1\right)=\dfrac{1}{4}\left(1-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow B=4\left|A\right|+\dfrac{1}{3^{100}}=1-\dfrac{1}{3^{100}}+\dfrac{1}{3^{100}}=1\)