Question

Chứng minh rằng:

A=1/3 + 2/3² + 3/3³ + 4/3³ + ....... +100/3¹⁰⁰ + 101/3¹⁰¹ < 3/4

Answer 1

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

\(3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\)

\(3A-A=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}+\frac{101}{3^{100}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}+\frac{101}{3^{101}}\right)\)

\(2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}-\frac{101}{3^{101}}\)

\(2A=1+\left(\frac{1-\frac{1}{3^{100}}}{2}\right)-\frac{101}{3^{101}}< 1+\frac{1}{2}=\frac{3}{2}\)

\(\Rightarrow A< \frac{3}{2}:2=\frac{3}{4}\)( đpcm )

Answer 2

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