Question

Find the value of ${\log }_{1003}(1+\frac{1}{2})+{\log }_{1003}(1+\frac{1}{3})+........+{\log }_{1003}(1+\frac{1}{2005})$

Answer 1

${\log }_{1003}\left(\frac{3}{2}\right)+{\log }_{1003}\left(\frac{4}{3}\right)+.....{\log }_{1003}\left(\frac{2006}{2005}\right)$

$={\log }_{1003}[\frac{3}{2}\times \frac{4}{3}\times \frac{5}{4}.....\frac{2005}{2004}\times \frac{2006}{2005}]$

$={\log }_{1003}[\frac{1}{2}\times \frac{2006}{1}]={\log }_{1003}1003=1$