Chapter F, Problem 47E

To determine

To prove: The formula for the sum of a finite geometric serious with first term a and common ratio $r\ne 1$, ${\displaystyle \sum _{i=1}^{n}a{r}^{i-1}=a+ar+a{r}^2+\cdots +a{r}^{n-1}}=\frac{a\left(r^n-1\right)}{r-1}$.

Explanation of Solution

Definition used:

If $a_m,a_{m+1},\mathrm{...},a_n$ are real numbers and m and n are integers such that $m\le n$, then ${\displaystyle \sum _{i=m}^n{a}_i=a_m+}{a}_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_n$.

Calculation:

By the above definition, the expression ${\displaystyle \sum _{i=1}^{n}a{r}^{i-1}}$ simplified as follows.

$\begin{array}{c}{\displaystyle \sum _{i=1}^{n}a{r}^{i-1}}=a{r}^{\left(1-1\right)}+a{r}^{\left(2-1\right)}+a{r}^{\left(3-1\right)}+\cdots +a{r}^{\left(n-1\right)}\\ =a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}\end{array}$

That is, ${\displaystyle \sum _{i=1}^{n}a{r}^{i-1}}=a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}$ (1)

Consider $S=a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}$ (2)

Multiply both side of the equation (2) by r gives,

$\begin{array}{c}rS=r\left(a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}\right)\\ =ar+a{r}^2+a{r}^3+\cdots +a{r}^n\end{array}$

That is, $rS=ar+a{r}^2+a{r}^3+\cdots +a{r}^n$ (3)

Subtract equation (2) from the equation (3).

$\begin{array}{c}rS-S=ar+a{r}^2+a{r}^3+\cdots +a{r}^n-\left(a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}\right)\\ S\left(r-1\right)=a{r}^n-a\\ S=\frac{a\left(r^n-1\right)}{r-1}\end{array}$

That is, $a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}=\frac{a\left(r^n-1\right)}{r-1}$ (4)

Compare the equation (1) with equation (4).

Thus, ${\displaystyle \sum _{i=1}^{n}a{r}^{i-1}}=a+a{r}^1+a{r}^2+\cdots +a{r}^{n-1}=\frac{a\left(r^n-1\right)}{r-1}$.

Hence the proof.