Question

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Answer 1

$\displaystyle lg\frac{1}{2} +lg\frac{2}{3} +lg\frac{3}{4} +...+lg\frac{99}{100} =lg\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot ... \cdot \frac{98}{99}\cdot \frac{99}{100} \right)=\\ \\ =lg\frac{1}{100} =lg100^{-1}=-lg100=-lg10^2=-2lg10=\mathbf{-2}$