Question

[Phạm Hà Quang Khánh]

C=1/3-2/3²+3/3³-4/3⁴+...+99/3⁹⁹-100/3^100<3/16

Answer 1

[_Never Give Up_ĐXRBBNBMCSNVCNĐTĐ_]

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

\(\Rightarrow3C=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow3C+C=4C=1-\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

Đặt A=\(1-\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)

\(\Rightarrow3A=3\times\left(1-\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)

\(\Rightarrow3A=3-1+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)

\(\Rightarrow3A+A=4A=3-\frac{1}{3^{99}}\)

\(\Rightarrow A=\frac{1}{4}\times\left(3-\frac{1}{3^{99}}\right)\)

Thay A vào ta có : \(4C=\frac{1}{4}\times\left(3-\frac{1}{3^{99}}\right)-\frac{1}{3^{100}}\)

\(C=\frac{1}{16}\times\left(3-\frac{1}{3^{99}}\right)-\frac{25}{3^{100}}\)

\(C=\frac{3}{16}-\frac{1}{16\times3^{99}}-\frac{25}{3^{100}}< \frac{3}{16}\)

Vậy C <\(\frac{3}{16}\)

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