Tính nhanh tổng A=1/2+1/4+1/8+1/16+1/32+1/64+1/128 28/09/2021 Bởi Adalynn Tính nhanh tổng A=1/2+1/4+1/8+1/16+1/32+1/64+1/128
Đáp án: \( \frac{{127}}{{128}}\) Giải thích các bước giải: $\begin{array}{l} A = \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}} \right)\\ \Rightarrow 2.A = 2.\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}} \right)\\ \,\,\,\,\,\,\,2.A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}}\\ \Rightarrow 2.A – A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} – \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}} \right)\\ \Rightarrow A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} – \frac{1}{2} – \frac{1}{4} – \frac{1}{8} – \frac{1}{{16}} – \frac{1}{{32}} – \frac{1}{{64}} – \frac{1}{{128}}\\ \Rightarrow A = 1 – \frac{1}{{128}} = \frac{{127}}{{128}}. \end{array}$ Bình luận
Đáp án: 127/128
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Giải thích các bước giả
Đáp án:
\( \frac{{127}}{{128}}\)
Giải thích các bước giải:
$\begin{array}{l}
A = \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}} \right)\\
\Rightarrow 2.A = 2.\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}} \right)\\
\,\,\,\,\,\,\,2.A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}}\\
\Rightarrow 2.A – A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} – \left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} + \frac{1}{{128}}} \right)\\
\Rightarrow A = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + \frac{1}{{64}} – \frac{1}{2} – \frac{1}{4} – \frac{1}{8} – \frac{1}{{16}} – \frac{1}{{32}} – \frac{1}{{64}} – \frac{1}{{128}}\\
\Rightarrow A = 1 – \frac{1}{{128}} = \frac{{127}}{{128}}.
\end{array}$