Toán cho A= -1,3 + 1/3^2 – 1/3^3 …….. + 1/3^50 – 1 /3^51 …. Tính a 09/09/2021 By Katherine cho A= -1,3 + 1/3^2 – 1/3^3 …….. + 1/3^50 – 1 /3^51 …. Tính a
Đáp án: Giải thích các bước giải: \[\begin{array}{l} A = – \frac{1}{3} + \frac{1}{{{3^2}}} – \frac{1}{{{3^3}}} + ….. + \frac{1}{{{3^{50}}}} – \frac{1}{{{3^{51}}}}\\ \Leftrightarrow 3A = – 1 + \frac{1}{3} – \frac{1}{{{3^2}}} + …. + \frac{1}{{{3^{49}}}} – \frac{1}{{{3^{50}}}}\\ \Leftrightarrow 3A + A = \left( { – 1 + \frac{1}{3} – \frac{1}{{{3^2}}} + …. + \frac{1}{{{3^{49}}}} – \frac{1}{{{3^{50}}}}} \right) + \left( { – \frac{1}{3} + \frac{1}{{{3^2}}} – \frac{1}{{{3^3}}} + …. + \frac{1}{{{3^{50}}}} – \frac{1}{{{3^{51}}}}} \right)\\ \Leftrightarrow 4A = – 1 – \frac{1}{{{3^{51}}}}\\ \Leftrightarrow 4A = – \frac{{{3^{51}} + 1}}{{{3^{51}}}} \Rightarrow A = – \frac{{{3^{51}} + 1}}{{{{4.3}^{51}}}} \end{array}\] Trả lời
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Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
A = – \frac{1}{3} + \frac{1}{{{3^2}}} – \frac{1}{{{3^3}}} + ….. + \frac{1}{{{3^{50}}}} – \frac{1}{{{3^{51}}}}\\
\Leftrightarrow 3A = – 1 + \frac{1}{3} – \frac{1}{{{3^2}}} + …. + \frac{1}{{{3^{49}}}} – \frac{1}{{{3^{50}}}}\\
\Leftrightarrow 3A + A = \left( { – 1 + \frac{1}{3} – \frac{1}{{{3^2}}} + …. + \frac{1}{{{3^{49}}}} – \frac{1}{{{3^{50}}}}} \right) + \left( { – \frac{1}{3} + \frac{1}{{{3^2}}} – \frac{1}{{{3^3}}} + …. + \frac{1}{{{3^{50}}}} – \frac{1}{{{3^{51}}}}} \right)\\
\Leftrightarrow 4A = – 1 – \frac{1}{{{3^{51}}}}\\
\Leftrightarrow 4A = – \frac{{{3^{51}} + 1}}{{{3^{51}}}} \Rightarrow A = – \frac{{{3^{51}} + 1}}{{{{4.3}^{51}}}}
\end{array}\]