Solutions

(a) Let us first prove the base case $n = 0$.

$$\begin{aligned} n = 0 \implies 1 &= \frac{1 - r^{0 + 1}}{1 - r} \\ &= \frac{1 - r}{1 - r} \\ &= 1 \end{aligned}$$

Moving on, we prove that the predicate holds for $n + 1$ if it does for $n$.

$$\begin{aligned} 1 + r + r^{2} + \dots + r^{n} + r^{n + 1} &= \frac{1 - r^{n + 1}}{1 - r} + r^{n + 1} \\ &= \frac{1 - r^{n + 1}}{1 - r} + \frac{(1 - r) r^{n + 1}}{1 - r} \\ &= \frac{1 - r^{n + 1} + r^{n + 1} - r^{n + 2}}{1 - r} \\ &= \frac{1 - r^{n + 2}}{1 - r} \end{aligned}$$

(b)

$$\begin{aligned} S (1 - r) &= S - Sr \\ &= 1 + r + \dots + r^{n} - (r + r^{2} + \dots + r^{n} + r^{n + 1}) \\ &= 1 - r^{n + 1} \\ & \implies S = \frac{1 - r^{n + 1}}{1 - r} \end{aligned}$$