Question

Tính C = \(\dfrac{1}{2}+\dfrac{1}{2^3}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{99}}\)

Answer 1

Lời giải:

Ta có \(C=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{99}}\)

\(\Rightarrow 4C=2+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}\)

\(\Rightarrow 4C-C=\left(2+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{97}}+\frac{1}{2^{99}}\right)\)

\(\Leftrightarrow 3C=2-\frac{1}{2^{99}}\)

\(\Leftrightarrow C=\frac{2}{3}-\frac{1}{3.2^{99}}\)