## Discrete Mathematics with Applications 4th Edition

Just substitute $d_m$ by the index, starting from $0$, because the sequence is defined for $m\geq1$, substitute the values and calculate.
$m \geq 0$ $\begin{split} d_m & = 1 + (\frac{1}{2})^m = 1 + \frac{1}{2^m}\\ & \\ d_0 & = 1 + (\frac{1}{2})^0 = 1 + \frac{1}{2^0} = 1 + \frac{1}{1} = 1 + 1 = 2 \\ & \\ d_1 & = 1 + (\frac{1}{2})^1 = 1 + \frac{1}{2^1} = 1 + \frac{1}{2} = \frac{3}{2} \\ & \\ d_2 & = 1 + (\frac{1}{2})^2 = 1 + \frac{1}{2^2} = 1 + \frac{1}{4} = \frac{5}{4} \\ & \\ d_3 & = 1 + (\frac{1}{2})^3 = 1 + \frac{1}{2^3} = 1 + \frac{1}{8} = \frac{9}{8} \\ & \\ \end{split}$