Problem 5

(a) Prove by induction on $n$ that: $$1 + r + r^2 + ... + r^n = \frac{1 - r^{n + 1}}{1 - r}$$ If $r \neq 1$.

Starting with $n = 0$: \[ \begin{align*} \frac{1 - r^1}{1 - r} &= 1 &= r^0 \\ \end{align*} \] Now, assuming $n = k$ show $n = k + 1$: \[ \begin{align*} 1 + r + ... + r^k + r^{k + 1} &= \frac{1 - r^{k + 1}}{1 - r} + r^{k + 1} \\ &= \frac{1 - r^{k + 1}}{1 - r} + \frac{r^{k + 1}(1 - r)}{1 - r} \\ &= \frac{1 - r^{k + 1}}{1 - r} + \frac{r^{k + 1} - r^{k + 2})}{1 - r} \\ &= \frac{1 - r^{k + 1} + r^{k + 1} - r^{k + 2}}{1 - r} \\ &= \frac{1 - r^{k + 2}}{1 - r} \\ &= \frac{1 - r^{(k + 1) + 1}}{1 - r} \\ \end{align*} \]

(b) Derive this result by setting $S = 1 + r + ... + r^n$, multiplying this equation by $r$, and solving the two equations for $S$. \[ \begin{align*} S &= 1 + r + ... + r^n \\ rS &= r + r^2 + ... + r^{n + 1} \\ \\ S(1 - r) &= S - rS \\ &= 1 - r^{n + 1} \\ \rightarrow S &= \frac{1 - r^{n + 1}}{1 - r} \\ \end{align*} \]

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