Chapter A2, Problem 2E

To determine

To show: $1+r+r^2+\mathrm{...}+r^n=\frac{1-r^{n+1}}{1-r}$ if $r\cancel{=}1$

Explanation of Solution

Proof:

The principle of mathematical induction states that if a certain mathematical relation is true for a base integer and, thereafter, assuming the relation to be true for an integer $k$ greater than the base integer, if we can prove the relation to be true for $k+1$ , the relation is true for all integers greater than the base integer.

Step 1: The formula holds for $n=1$ because

  $1+r=\frac{\left(1+r\right)\left(1-r\right)}{\left(1-r\right)}=\frac{1-r^2}{1-r}$

Step 2: Suppose $1+r+r^2+\mathrm{...}+r^k=\frac{1-r^{k+1}}{1-r}$

Then

  $\begin{array}{l}1+r+r^2+\mathrm{...}+r^{k+1}=\frac{1-r^{k+1}}{1-r}+r^{k+1}\\ =\frac{1-r^{k+2}}{1-r}\end{array}$

Hence, the principal of mathematical induction guarantees this is true for all $n$ .

Conclusion:

The principal of mathematical induction guarantees this is true for all $n$ .