Geometric Series

Adding Monthly Batches of Seedlings

Suppose a nursery prepares a new batch of seedlings each month according to the following idealized pattern:

Month	January	February	March	April	May
New seedlings prepared	$4$	$12$	$36$	$108$

The number prepared in each month forms a geometric sequence. Adding the new batches from the first month through a chosen month forms a geometric series. Because every batch is newly prepared, this sum does not count the same seedlings twice.

A geometric series is the sum of the terms of a geometric sequence. In that sequence, each term is obtained by multiplying the preceding term by a constant factor $r$ . When the preceding term is nonzero, this factor is also the quotient of the two consecutive terms.

In the seedling data:

The first term ( $a$ or $U_1$ ) is $4$ .
The ratio $(r) = \frac{12}{4} = \frac{36}{12} = \frac{108}{36} = \frac{324}{108} = 3$ .

So, the sequence of monthly batches is $4, 12, 36, 108, 324$ . The geometric series is the sum of the terms of this sequence:

Sum of the first $2 \text{ months}$ $(S_2) = 4 + 12 = 16$
Sum of the first $3 \text{ months}$

Finding the Formula for the Sum of the First Terms

How can we calculate the sum of the first $n$ terms $(S_n)$ without adding them one by one? The following calculation derives the formula.

Consider this table which shows the process of rediscovering the formula for the sum of a geometric series:

Notation	Direct Summation	Using $U_{n+1}$ and $U_1$

For $r\ne1$ , the bottom-right cell gives the general form:

S_n = \frac{U_{n+1} - U_1}{r - 1}

We know that the formula for the $n$ -th term in a geometric sequence is $U_n = ar^{n-1}$ . Thus, $U_{n+1} = ar^{(n+1)-1} = ar^n$ . Substitute and into the formula:

S_n = \frac{ar^n - a}{r - 1}

This is the formula for the sum of the first $n$ terms of a geometric series.

We can also derive the same result by multiplying the sum by $r$ and subtracting:

S_n=a+ar+ar^2+\cdots+ar^{n-1}

The middle terms cancel. This explains why the formula contains only the first term and the next power after the final term.

Geometric Series Formula

In general, the formula to calculate the sum of the first $n$ terms of a geometric series is:

S_n=\frac{a(r^n-1)}{r-1},\quad r\ne1

An equivalent form is

S_n=\frac{a(1-r^n)}{1-r},\quad r\ne1

Both forms work for positive or negative ratios as long as $r\ne1$ . If $r=1$ , all terms equal $a$ , so:

S_n=na

Legend:

$S_n$ = sum of the first $n$ terms
$a$ = first term $(U_1)$

Calculating Total Bicycle Production

In a simplified production plan for $2020$ , a bicycle company's newly produced units each month form a geometric sequence. January production is $120 \text{ units}$ , and April production is $3{,}240 \text{ units}$ . What is the total production from January through May?

Solution:

January Production $(U_1) = a = 120$
April Production $(U_4) = 3{,}240$

Step 1: Find the ratio $(r)$

U_4 = ar^{4-1} = ar^3

Step 2: Calculate $S_5$ . Since $r = 3 > 1$ , use the formula:

S_n = \frac{a(r^n - 1)}{r - 1}

The total bicycle production up to May is $14{,}520 \text{ units}$ .

The addition is meaningful here because each monthly value represents newly produced bicycles. Adding repeated measurements of one unchanged stock would instead count the same objects more than once.

Practice with a Negative Ratio

Calculate the finite sum:

81-27+9-3+1

The first term is $a=81$ , the ratio is $r=-\frac13$ , and there are $n=5$ terms. Therefore:

S_5=81\left(\frac{1-(-1/3)^5}{1-(-1/3)}\right)

Check by direct addition: $81-27+9-3+1=61$ .

$S_2$	$4 + 12 = 16$	$S_2 = \frac{36 - 4}{3 - 1} = \frac{32}{2} = 16$	$S_2 = \frac{U_3 - U_1}{r - 1}$
$S_3$	$4 + 12 + 36 = 52$
$S_4$	$4 + 12 + 36 + 108 = 160$
$\vdots$	$\dots$	$\dots$	$\dots$
$S_n$	$\dots$	$\dots$	$S_n = \frac{U_{n+1} - U_1}{r - 1}$