cho D=1/3 + 1/3^2 + 1/3^3 +…+1/3^100. So sánh D với 1/2 10/10/2021 Bởi Eden cho D=1/3 + 1/3^2 + 1/3^3 +…+1/3^100. So sánh D với 1/2
Đáp án: $\begin{array}{l}D = \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + … + \dfrac{1}{{{3^{100}}}}\\ \Rightarrow 3.D = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{3^{99}}}}\\ \Rightarrow 3D – D = 2D = 1 – \dfrac{1}{{{3^{100}}}}\\ \Rightarrow D = \dfrac{1}{2} – \dfrac{1}{{{{2.3}^{100}}}} < \dfrac{1}{2}\\Vậy\,D < \dfrac{1}{2}\end{array}$ Bình luận
Ta có: ` D = 1/3 + \frac{1}{3^{2}} + \frac{1}{3^{3}} + … + \frac{1}{3^{100}} ` ` <=> 3D = 1 + 1/3 + \frac{1}{3^{2}} + … + \frac{3^{99}} ` ` <=> 3D – D = ( 1 + 1/3 + \frac{1}{3^{2}} + … + \frac{1}{3^{99}} ) – ( 1/3 + \frac{1}{3^{2}} + \frac{1}{3^{3}} + … + \frac{1}{3^{100}} ) ` ` <=> 2D = 1 + 1/3 + \frac{1}{3^{2}} + … + \frac{3^{99}} – 1/3 – \frac{1}{3^{2}} – \frac{1}{3^{3}} – … – \frac{1}{3^{100}} ` ` <=> 2D = 1 + (1/3 – 1/3) + ( \frac{1}{3^{2}} – \frac{1}{3^{2}} ) + … + (\frac{1}{3^{99}} – \frac{1}{3^{99}}) – \frac{1}{3^{100}} ` ` <=> 2D = 1 – \frac{1}{3^{100}} ` ` <=> D = \frac{1 – \frac{1}{3^{100}}}{2} ` ` <=> D = 1/2 – \frac{1}{2.3^{100}} ` Do: ` 1/2 – \frac{1}{2.3^{100}} < 1/2 ` ` => D < 1/2 ` ` (đpcm) ` Bình luận
Đáp án:
$\begin{array}{l}
D = \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + … + \dfrac{1}{{{3^{100}}}}\\
\Rightarrow 3.D = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + … + \dfrac{1}{{{3^{99}}}}\\
\Rightarrow 3D – D = 2D = 1 – \dfrac{1}{{{3^{100}}}}\\
\Rightarrow D = \dfrac{1}{2} – \dfrac{1}{{{{2.3}^{100}}}} < \dfrac{1}{2}\\
Vậy\,D < \dfrac{1}{2}
\end{array}$
Ta có:
` D = 1/3 + \frac{1}{3^{2}} + \frac{1}{3^{3}} + … + \frac{1}{3^{100}} `
` <=> 3D = 1 + 1/3 + \frac{1}{3^{2}} + … + \frac{3^{99}} `
` <=> 3D – D = ( 1 + 1/3 + \frac{1}{3^{2}} + … + \frac{1}{3^{99}} ) – ( 1/3 + \frac{1}{3^{2}} + \frac{1}{3^{3}} + … + \frac{1}{3^{100}} ) `
` <=> 2D = 1 + 1/3 + \frac{1}{3^{2}} + … + \frac{3^{99}} – 1/3 – \frac{1}{3^{2}} – \frac{1}{3^{3}} – … – \frac{1}{3^{100}} `
` <=> 2D = 1 + (1/3 – 1/3) + ( \frac{1}{3^{2}} – \frac{1}{3^{2}} ) + … + (\frac{1}{3^{99}} – \frac{1}{3^{99}}) – \frac{1}{3^{100}} `
` <=> 2D = 1 – \frac{1}{3^{100}} `
` <=> D = \frac{1 – \frac{1}{3^{100}}}{2} `
` <=> D = 1/2 – \frac{1}{2.3^{100}} `
Do: ` 1/2 – \frac{1}{2.3^{100}} < 1/2 `
` => D < 1/2 ` ` (đpcm) `