MCR3U Geometric Series: Notes and Practice

A legend tells of a ruler who offered a reward of rice on a chessboard: 1 grain on the first square, 2 on the second, 4 on the third, doubling every square. The ruler agreed, thinking it modest, but the total on all 64 squares is more than 18 quintillion grains, far more rice than has ever been grown. Adding a geometric sequence is a geometric SERIES, and because each term is a multiple of the last, these sums can explode (or, when the ratio is less than 1, settle down) in ways our intuition misjudges. Phone trees, the spread of a rumour, the total distance travelled by a bouncing ball and, later in this unit, the value of a savings plan are all geometric series.

Deriving the formula

Gauss's pairing trick relied on the first-plus-last symmetry of an arithmetic sequence, which a geometric sequence doesn't have: in $1 + 2 + 4 + 8$ , the pairs $1 + 8$ and $2 + 4$ are not equal. Instead, geometric series get summed with a shift-and-subtract trick. Write out $S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}$ . Now multiply every term by $r$ : $r S_{n} = a r + a r^{2} + a r^{3} + \dots + a r^{n}$ . Multiplying by $r$ shifts every term one place along, so the two lists share almost every term. Subtracting cancels everything except the very first term of $S_{n}$ and the very last term of $r S_{n}$ .

S_{n} r S_{n} S_{n} - r S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1} = a + a r + a r^{2} + \dots + a r^{n - 1} + a r^{n} = a - a r^{n}

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

Both forms give the same value; use whichever avoids a negative denominator.

The two forms are the same fraction with the top and bottom both multiplied by $- 1$ . When $r > 1$ , use $\frac{a ( r ^{n} - 1 )}{r - 1}$ so that everything stays positive; when $0 < r < 1$ , use $\frac{a ( 1 - r ^{n} )}{1 - r}$ . Notice the exponent is $n$ , the NUMBER of terms, not $n - 1$ : the $a r^{n}$ in the formula is the term just past the end of the sum. A quick check shows it works: $2 + 6 + 18 + 54 + 162 = 242$ , and the formula gives $\frac{2 ( 3 ^{5} - 1 )}{3 - 1} = \frac{2 ( 242 )}{2} = 242$ .

Fig 1Each new term is a shrinking or growing slice. When the ratio is below 1, the slices get so small that the running total barely changes.

The special case r = 1

r = 1

, the formula above involves dividing by

1 - r = 0

, which is undefined. But if

r = 1

every term already equals

a

, so the sum is simply

S_{n} = na

. The derivation itself shows why: dividing by

1 - r

was the last step, and it is only allowed when

1 - r \neq = 0

. A negative ratio causes no trouble at all; for

r = - 2

the denominator is

1 - (- 2) = 3

, and the powers

r^{n}

simply alternate in sign, so an alternating series can have a sum smaller than its first term.

VideoEddie Woo multiplies the series by r, subtracts, and watches almost every term cancel, which is the same derivation this lesson uses for the geometric series formula.Watch on YouTube

Finding n before you sum

In MCR3U, the exponent $n$ (the number of terms) is what usually needs the most care. Often a question shows the last term instead of telling you $n$ , as in $3 + 6 + 12 + \dots + 3072$ . Find $n$ first by solving $t_{n} = a r^{n - 1}$ : $3 (2)^{n - 1} = 3072$ gives $2^{n - 1} = 1024 = 2^{10}$ , so $n = 11$ . Then $S_{11} = \frac{3 ( 2 ^{11} - 1 )}{2 - 1} = 3 (2047) = 6141 .$ A shortcut skips $n$ entirely: since $a r^{n} = r \cdot t_{n}$ , the sum formula can be written using the last term directly.

S_{n} = \frac{r t _{n} - a}{r - 1} e.g. \frac{2 ( 3072 ) - 3}{2 - 1} = 6141

Series	a	r	n	Sum
$1 + 3 + 9 + 27$	1	3	4	$\frac{3 ^{4} - 1}{2} = 40$
$64 + 32 + \dots + 2$	64	0.5	6	$\frac{64 ( 1 - 0. 5 ^{6} )}{0.5} = 126$
$4 - 8 + 16 - 32$	4	−2	4	$\frac{4 ( 16 - 1 )}{- 3} = - 20$
$5 + 5 + 5 + 5 + 5 + 5$	5	1	6	$6 (5) = 30$

Applications and recognising a geometric series

A phone tree where each person calls 3 others reaches $1 + 3 + 9 + \dots + 3^{5} = \frac{3 ^{6} - 1}{2} = 364$ people in 6 levels. A ball dropped from 10 m that rebounds to 60% of its height travels 10 m down, then UP and DOWN each rebound height: $6, 3.6, 2.16, \dots$ By the time it hits the ground for the 5th time it has travelled $10 + 2 (6 + 3.6 + 2.16 + 1.296) = 10 + 2 (13.056) = 36.112$ m. The bracket is a geometric series with $a = 6$ , $r = 0.6$ , $n = 4$ . The two traps are forgetting the factor of 2 (each rebound is travelled twice) and adding the initial drop into the doubled part.

AArithmetic series

Consecutive terms have a common DIFFERENCE $d$
Derived by writing the sum forwards and backwards (pairing)
$S_{n} = \frac{n}{2} (a + t_{n})$
Grows roughly like $n^{2}$

BGeometric series

Consecutive terms have a common RATIO $r$
Derived by multiplying by $r$ and subtracting ( $S_{n} - r S_{n}$ )
$S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$ , $r \neq = 1$
Explodes when $∣ r ∣ > 1$ ; levels off toward a limit when $∣ r ∣ < 1$

Confirm the terms share a common ratio: divide at least two consecutive pairs. Convert percentages to a ratio (a 60% rebound means $r = 0.6$ ; a 5% increase means $r = 1.05$ ).
Identify $a$ as the first term actually being added. In a bounce problem the first rebound is $10 (0.6) = 6$ , not the 10 m drop.
Find $n$ , the number of terms. If only the last term is given, solve $a r^{n - 1} = t_{n}$ by matching bases, or use the last-term form of the formula.
Choose the form of the formula that keeps the numbers positive, substitute, and keep full calculator precision until the final answer.
Sanity check: for $r > 1$ the sum should be larger than the last term (with $r = 2$ it is just under double the last term, as in $6141$ vs $3072$ ); for $0 < r < 1$ the sum should be less than $\frac{a}{1 - r}$ (for $6 + 3 + 1.5 + \dots$ that ceiling is 12).

Geometric Series

Key things to remember

Deriving the formula

Finding n before you sum

Applications and recognising a geometric series

Worked examples

Where marks get lost

Flashcards 8

Practice