Question

[Nguyễn Văn Vi Duy Hưng]

Cho E=\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}.\)Chứng minh:C<\(\dfrac{5}{3}\)

Answer 1

[Izuku Midoriya(Deku) (Team Friday N...]

$C=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}$

$3C=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}$

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

$2C=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}$

$6C=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}$

$6C-2C=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)$

$4C=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}$

$4C=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}$

$4C=3-\frac{203}{3^{100}}<3$

$\Rightarrow C<\frac{3}{4}\left(đpcm\right)$

Answer 2

[Izuku Midoriya(Deku) (Team Friday N...]

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