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Let $P(n)$ be the statement $$ P(n) : \sum\limits^{n}_{i=0}r^i = \dfrac{1-r^{1+n}}{1-r}\text{ for all }n \in \mathbb{N}\text{.} $$ I am stuck at the base case: $$P(1):1 + r = \dfrac{1-r^2}{1-r}\text{.}$$

I am stuck as to how I can show $P(1)$ is true.

Daniel W. Farlow
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Guest
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    Hint: $a^2-b^2=(a+b)(a-b)$ –  Aug 14 '14 at 01:42
  • Base Case is P(0) not P(1) – user137481 Aug 14 '14 at 01:42
  • @user137481 Natural Numbers start at 1 ... so should the base case not be P(1) – Guest Aug 14 '14 at 01:43
  • @user14982305 Thanks I was able to complete the proof ... Careless mistakes always get the best of me in proofs :P – Guest Aug 14 '14 at 01:46
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    That depends on your teacher. In some courses, Natural Numbers start at 0 – user137481 Aug 14 '14 at 01:47
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    Note that the problem really should have specified that $r\ne 1$. As to the base case, it depends on the exact statement of the problem. If we are asked to prove the result is true for every natural number $n$, then the base case is $n=1$. If we are asked to prove the result holds for every non-negative integer, then the base case is $n=0$. – André Nicolas Aug 14 '14 at 01:48

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Note: First note that you are really summing the geometric series ($a,r\in\mathbb{R}, a\neq 0, r\neq 1$) $$ a+ar+ar^2+\cdots+ar^n=a\frac{r^{n+1}-1}{r-1}=a\frac{1-r^{n+1}}{1-r} $$ with $a=1$. Let's tidy up the question wording just a little bit and then prove your claim.

Claim: Let $a$ and $r$ be real numbers with $a\neq 0$ and $r\neq 1$. Prove that for each integer $n\geq 1$ that $$ a+ar+ar^2+\cdots+ar^n=a\frac{r^{n+1}-1}{r-1}. $$ Proof. Fix $r\in\mathbb{R}, r\neq 1$, and let $S(n)$ denote the following statement for $n\geq 1$: $$ S(n) : 1+r+r^2+\cdots+r^n=\frac{r^{n+1}-1}{r-1}. $$ In order to solve this problem, it is sufficient for us to prove that $S(n)$ holds for each $n\geq 1$; that is, it suffices to consider only $a=1$. For $r=0$, we can see that $S(n)$ becomes $1=1$. Thus, assume that $r\neq 0$ without loss of generality.

Base step ($n=1$): The statement $S(1)$ says that $1+r=\frac{r^2-1}{r-1}$, which is true since $r^2-1=(r+1)(r-1)$ and $r\neq 1$. [Also observe that you could begin the induction at $n=0$ because $S(0)$ simply says $1=\frac{r-1}{r-1}$.]

Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and suppose that $S(k)$ is true; that is, assume that $$ S(k) : 1+r+r^2+\cdots+r^k=\frac{r^{k+1}-1}{r-1} $$ is true. We must show that $S(k+1)$ follows where $$ S(k+1) : 1+r+r^2+\cdots+r^k+r^{k+1}=\frac{r^{k+2}-1}{r-1}. $$ Starting with the left-hand side of $S(k+1)$, \begin{align} \text{LHS} &= \color{red}{1+r+r^2+\cdots+r^k}+r^{k+1}\tag{by definition}\\[1em] &= \color{red}{\frac{r^{k+1}-1}{r-1}}+r^{k+1}\tag{by $S(k)$}\\[1em] &= \frac{r^{k+1}-1}{r-1}+\frac{r^{k+1}(r-1)}{r-1}\tag{common denom.}\\[1em] &= \frac{r^{k+1}-1+r^{k+1}(r-1)}{r-1}\tag{combine like terms}\\[1em] &= \frac{r^{k+1}-1+r^{k+2}-r^{k+1}}{r-1}\tag{expand}\\[1em] &= \frac{r^{k+2}-1}{r-1},\tag{simplify} \end{align} we see that the right-hand side of $S(k+1)$ follows. Thus, $S(k)\to S(k+1)$, completing the inductive step.

By mathematical induction, for all $n\geq 1, S(n)$ holds true. $\blacksquare$

Last note: There are many identities that use this main result. For example, a question was posed not long at all ago to prove that $\sum_{i=0}^n 2^i=2^{n+1}-1$. One can easily set $r=2$ in what we just proved to see that this result is true.

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Daniel W. Farlow
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