?

Number Theory

Free Version

Upgrade subject to access all content

Easy

Finite Geometric Series and Induction

NUMTH-231BT9

Which of the following proofs by induction are correct?

A

Let $P(n)$ be the property that $$\sum_{k = 1}^{n}r^k = \frac{r - r^{n + 1}}{1 - r}$$ where $r \neq 1$. Suppose that for $m\in \mathbb{N}$ that $P(m)$ is true. We want to show that $P(m + 1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{m + 1}r^k &= \sum_{k = 1}^{m}r^k + r^{m + 1}\\\ &= \frac{r - r^{m + 1}}{1 - r} + r^{m + 1} \\\ &= \frac{r - r^{m + 1} + r^{m + 1}(1 - r)}{1 - r}\\\ &= \frac{r - r^{m + 2}}{1 - r}.\end{align}$$ Thus, $P(m + 1)$ is true and we are done.

B

Let $P(n)$ be the property that $$\sum_{k = 1}^{n}r^k = \frac{r - r^{n + 1}}{1 - r}$$ First, we will show that $P(1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{1} r^k &= r \\\ &= \frac{r(1 - r)}{1 - r}\\\ &= \frac{r - r^2}{1 - r}.\end{align}$$ Hence, $P(1)$ is true. Suppose that for $m\in \mathbb{N}$ that $P(m)$ is true. We want to show that $P(m + 1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{m + 1}r^k &= \sum_{k = 1}^{m}r^k + r^{m + 1}\\\ &= \frac{r - r^{m + 1}}{1 - r} + r^{m + 1} \\\ &= \frac{r - r^{m + 1} + r^{m + 1}(1 - r)}{1 - r}\\\ &= \frac{r - r^{m + 2}}{1 - r}.\end{align}$$ Thus, $P(m + 1)$ is true and we are done.

C

Let $P(n)$ be the property that $$\sum_{k = 1}^{n}r^k = \frac{r - r^{n + 1}}{1 - r}$$ where $r \neq 1$. First, we will show that $P(1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{1} r^k &= r \\\ &= \frac{r(1 - r)}{1 - r}\\\ &= \frac{r - r^2}{1 - r}.\end{align}$$ Hence, $P(1)$ is true. Suppose that for $m\in \mathbb{N}$ that $P(m)$ is true. We want to show that $P(m + 1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{m + 1}r^k &= \sum_{k = 1}^{m}r^k + r^{m + 1}\\\ &= \frac{r - r^{m + 1}}{1 - r} + r^{m + 1} \\\ &= \frac{r - r^{m + 1} + r^{m + 1}(1 - r)}{1 - r}\\\ &= \frac{r - r^{m + 2}}{1 - r}.\end{align}$$ Thus, $P(m + 1)$ is true and we are done.

D

Let $P(n)$ be the property that $$\sum_{k = 1}^{n}r^k = \frac{r - r^{n + 1}}{1 - r}$$ where $r \neq 1$. First, we will show that $P(1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{1} r^k &= r \\\ &= \frac{r(1 - r)}{1 - r}\\\ &= \frac{r - r^2}{1 - r}.\end{align}$$ Hence, $P(1)$ is true. Suppose that for $m\in \mathbb{N}$ that $P(m)$ is true. We want to show that $P(m + 1)$ is true. Note that $$\begin{align} \sum_{k = 1}^{m + 1}r^k &= \sum_{k = 1}^{m}r^k + r^{m + 1}\\\ &= \frac{r - r^m}{1 - r} + r^{m + 1} \\\ &= \frac{r - r^m + r^{m + 1}(1 - r)}{1 - r}\\\ &= \frac{r - r^m + r^{m + 1} - r^{m + 2}}{1 - r}.\end{align}$$ Thus, $P(m + 1)$ is false and the proposition is false.