Question

Tính ta được

Answer 1

\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow2B=1+\frac{1}{2}+...+\left(\frac{1}{2}\right)^{98}\)

\(\Rightarrow2B-B=\left[1+\frac{1}{2}+...+\left(\frac{1}{2}\right)^{98}\right]-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{99}\right]\)

\(\Rightarrow B=1-\left(\frac{1}{2}\right)^{99}\)

Vậy \(B=1-\left(\frac{1}{2}\right)^{99}\)

Answer 2

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