A = $\frac{1}{3}$ + $\frac{2}{3^2}$ + $\frac{3}{3^3}$ + $\frac{4}{3^4}$ +…+ $\frac{100}{3 mũ 100}$ < $\frac{3}{4}$ (Gợi ý tính 3A, đặt rồi tính B)
A = $\frac{1}{3}$ + $\frac{2}{3^2}$ + $\frac{3}{3^3}$ + $\frac{4}{3^4}$ +…+ $\frac{100}{3 mũ 100}$ < $\frac{3}{4}$ (Gợi ý tính 3A, đặt rồi tính B)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{1}{3} + \dfrac{2}{{{3^2}}} + \dfrac{3}{{{3^3}}} + \dfrac{4}{{{3^4}}} + ….. + \dfrac{{100}}{{{3^{100}}}}\\
\Rightarrow 3A = 3.\dfrac{1}{3} + 3.\dfrac{2}{{{3^2}}} + 3.\dfrac{3}{{{3^3}}} + 3.\dfrac{4}{{{3^4}}} + ….. + 3.\dfrac{{100}}{{{3^{100}}}}\\
\Leftrightarrow 3A = 1 + \dfrac{2}{3} + \dfrac{3}{{{3^2}}} + \dfrac{4}{{{3^3}}} + …… + \dfrac{{100}}{{{3^{99}}}}\\
\Rightarrow 3A – A = \left( {1 + \dfrac{2}{3} + \dfrac{3}{{{3^2}}} + \dfrac{4}{{{3^3}}} + …… + \dfrac{{100}}{{{3^{99}}}}} \right) – \left( {\dfrac{1}{3} + \dfrac{2}{{{3^2}}} + \dfrac{3}{{{3^3}}} + \dfrac{4}{{{3^4}}} + ….. + \dfrac{{100}}{{{3^{100}}}}} \right)\\
\Leftrightarrow 2A = 1 + \left( {\dfrac{2}{3} – \dfrac{1}{3}} \right) + \left( {\dfrac{3}{{{3^2}}} – \dfrac{2}{{{3^2}}}} \right) + \left( {\dfrac{4}{{{3^3}}} – \dfrac{3}{{{3^3}}}} \right) + …. + \left( {\dfrac{{100}}{{{3^{99}}}} – \dfrac{{99}}{{{3^{99}}}}} \right) – \dfrac{{100}}{{{3^{100}}}}\\
\Leftrightarrow 2A = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + ….. + \dfrac{1}{{{3^{99}}}} – \dfrac{{100}}{{{3^{100}}}}\\
B = 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + ….. + \dfrac{1}{{{3^{99}}}}\\
\Leftrightarrow 3B = 3 + 3.\dfrac{1}{3} + 3.\dfrac{1}{{{3^2}}} + 3.\dfrac{1}{{{3^3}}} + 3.\dfrac{1}{{{3^4}}} + …. + 3.\dfrac{1}{{{3^{99}}}}\\
\Leftrightarrow 3B = 3 + 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + …. + \dfrac{1}{{{3^{98}}}}\\
\Rightarrow 3B – B = \left( {3 + 1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + …. + \dfrac{1}{{{3^{98}}}}} \right) – \left( {1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}} + \dfrac{1}{{{3^4}}} + ….. + \dfrac{1}{{{3^{99}}}}} \right)\\
\Leftrightarrow 2B = 3 – \dfrac{1}{{{3^{99}}}}\\
\Rightarrow 2A = B – \dfrac{{100}}{{{3^{100}}}} < B\\
\Rightarrow A = \dfrac{B}{2} = \dfrac{{3 – \dfrac{1}{{{3^{99}}}}}}{4} = \dfrac{3}{4} – \dfrac{1}{{{{4.3}^{99}}}} < \dfrac{3}{4}
\end{array}\)
A=$\frac{1}{3}$+$\frac{2}{3^{2}}$+$\frac{3}{3^{3}}$+$\frac{4}{3^{4}}$…+$\frac{100}{3^{100}}$
3A=1+$\frac{2}{3}$+$\frac{3}{3^{2}}$+$\frac{4}{3^{3}}$+…+$\frac{100}{3^{99}}$
3A-2A=1+$\frac{-1}{3^{2}}$+$\frac{-2}{3^{3}}$+…+$\frac{-98}{3^{99}}$-$\frac{200}{3^{100}}$
A=1-($\frac{1}{3^{2}}$+$\frac{2}{3^{3}}$+…+$\frac{98}{3^{99}}$+$\frac{100}{3^{100}}$)-$\frac{100}{3^{100}}$
A=1-A-$\frac{100}{3^{100}}$
A+A=1-$\frac{100}{3^{100}}$
2A=1-$\frac{100}{3^{100}}$
A=$\frac{1-\frac{100}{3^{100}}}{2}$
A=$\frac{2}{4}$ -$\frac{50}{3^{100}}$<$\frac{3}{4}$
Vậy $\frac{1}{3}$+$\frac{2}{3^{2}}$+$\frac{3}{3^{3}}$+$\frac{4}{3^{4}}$…+$\frac{100}{3^{100}}$<$\frac{3}{4}$(ĐPCM)
Cho mik xin TRẢ LỜI HAY NHẤT NHÉ!!!