Ta có: \(C=\frac12+\frac{1}{2^3}+\cdots+\frac{1}{2^{99}}\)
=>\(4C=2+\frac12+\cdots+\frac{1}{2^{97}}\)
=>4C-C=\(2+\frac12+\cdots+\frac{1}{2^{97}}-\frac12-\frac{1}{2^3}-\cdots-\frac{1}{2^{99}}\)
=>3C=\(2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\)
=>\(C=\frac{2^{100}-1}{3\cdot2^{99}}\)
