Question

\(A=\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)+\(\frac{1}{64}\)+\(\frac{1}{128}\)+\(\frac{1}{256}\)

Tính A?

Answer 1

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{128}-\frac{1}{256}\)

\(A=1-\frac{1}{256}\)

\(A=\frac{255}{256}\)

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Answer 2

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}...-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)

\(A=1-0+0+0+...+0+0-\frac{1}{256}\)

\(A=1-\frac{1}{256}\)

\(A=\frac{255}{256}\)