Problem 3

$\sum_{r=2}^{63} \log_2 \left( 1 + \dfrac1r \right) = \log_2(\prod_{r=2}^{63} (1 + \dfrac1r)) = \log_2(\dfrac 32 \cdot \dfrac 43 \cdot \dfrac 54 \cdot \ldots \cdot \dfrac {64}{63}) =$ $\log_2 \frac {64}{2} = \log_2 32 = 5$

$log_2(1+\frac{1}{r}) = log_2(\frac{r+1}{r}) = log_2(r+1)-log_2(r)$ From here we just put sigma into it and the solution will be $(log_2 64-log_2 63)+(log_2 63-log_2 62)+\dots+(log_2 3-log_2 2) = log_2 64 - log_2 2 = 6-1 = 5$