Algebra 2 · A.SSE.B.4

Geometric series

The terms 5, 15, 45, and onward triple each time. What do the first 8 of them add up to, and is there a way to know without adding them one by one?

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Geometric sequences and the explicit formulaAlgebra 1 Sigma notationAlgebra 2

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Infinite geometric seriesAlgebra 2

Where you are 10 in Sequences and series

Algebra9

Sequences and series10

▴ 5 above Sigma notation Arithmetic series and partial sums

Geometric series

Recursive formulas Infinite geometric series

When you reach for it

The sequence 5, 15, 45, 135, \ldots triples at every step. It is geometric, with first term a_1 = 5 and common ratio r = 3, so its nth term is 5 \cdot 3^{n-1}. Its total is another matter. The first 8 terms are

5 + 15 + 45 + 135 + 405 + 1215 + 3645 + 10935

and adding them one at a time takes 7 additions on numbers that grow fast. For 30 terms the last one alone has fifteen digits.

Reach for this skill whenever the thing wanted is a total of terms that each multiply the one before by the same number. Totals like that turn up wherever growth is by a percentage: a salary that rises by 2.5\% a year, savings deposits that each earn interest for a different number of years, the payments that clear a loan. They also turn up where something shrinks by a fixed fraction, like the heights of a bouncing ball. In sigma notation such a total is \sum_{k=1}^{n} a_1 r^{k-1}, and the move below turns it into one short formula.

A sum of the terms of a geometric sequence is a geometric series. As with the arithmetic series, S_n stands for the sum of the first n terms.

The procedure

Multiply by the ratio and subtract

Call the total of the first 8 terms S. Multiplying every term by 3 gives the very next term of the sequence, so 3S is the same list moved one place along:

\begin{aligned} S &= 5 + 15 + 45 + \cdots + 10935 \\ 3S &= \phantom{5 + {}} 15 + 45 + \cdots + 10935 + 32805 \end{aligned}

The two lines share every term from 15 to 10935. Subtracting the first line from the second removes all of them, and only the two ends survive: the 32805 that 3S has and S does not, and the 5 that S has and 3S does not. That is

3S - S = 32805 - 5

so 2S = 32800 and S = 16400. Seven additions have become one subtraction and one division.

The same move works for any geometric series. Write out the first n terms, multiply through by r (the distributive property), and line the two up:

\begin{aligned} S_n &= a_1 + a_1 r + \cdots + a_1 r^{n-1} \\ rS_n &= a_1 r + \cdots + a_1 r^{n-1} + a_1 r^n \end{aligned}

Subtracting the second line from the first, every term from a_1 r to a_1 r^{n-1} appears once with each sign and cancels. What is left is

S_n - rS_n = a_1 - a_1 r^n

Both sides factor, S_n(1 - r) = a_1(1 - r^n), and as long as r \ne 1 the factor 1 - r is not zero, so both sides may be divided by it.

The formula

For a geometric sequence with first term a_1 and common ratio r \ne 1, the sum of the first n terms is

S_n = \frac{a_1(1 - r^n)}{1 - r}

Multiplying the top and the bottom by -1 gives the same number in a second form,

S_n = \frac{a_1(r^n - 1)}{r - 1}

which keeps both brackets positive when r > 1. Use whichever is tidier. When r = 1 every term is a_1 and the sum is simply n a_1; the formula does not apply there because it would divide by zero.

Since a_1 r^n = r \cdot a_1 r^{n-1} = r \, a_n, the formula can also be written with the last term in place of n:

S_n = \frac{a_1 - r\,a_n}{1 - r}

which is handy when the sum is given as a list from a first term to a last one.

The formula needs three numbers. The first term a_1 is the first term actually in the sum. The ratio r is any term divided by the one before it, and it must be the same all the way along. The number of terms n is a count, not the last exponent: in \sum_{k=1}^{n} a_1 r^{k-1} the exponents run from 0 to n - 1, which is n terms.

Figure 1 shows the case a_1 = 1, r = 2. The formula gives 1 + 2 + 4 + \cdots + 2^{n-1} = 2^n - 1, and the picture shows why: the stacked terms always fall exactly one short of the next one.

Figure 1The blocks 1, 2, 4, 8, 16 stacked together reach 31, one short of the next term, 32. Every power of 2 is one more than all the powers before it put together.

Deposits and repayments

Suppose an equal amount P is paid into an account at the end of every period, and the account grows by the factor g each period (so g = 1.046 for 4.6\% a period). Look at the account right after the nth deposit. The deposit just made has earned nothing and is worth P. The one before has grown for one period and is worth Pg. The very first has grown for n - 1 periods and is worth Pg^{n-1}. The balance is

\begin{aligned} &P + Pg + Pg^2 + \cdots + Pg^{n-1} \\ &\quad = \frac{P(g^n - 1)}{g - 1} \end{aligned}

a geometric series with first term P, ratio g and n terms. The ratio is the growth factor, not the percentage: at 4.6\% it is 1.046, never 0.046.

A loan repaid in equal installments is the same series seen from the lender’s side, as Example 7 shows.

Worked examples

Example 1

The first eight terms, and then twelve

Find the sum of the first 8 terms of 5 + 15 + 45 + \cdots, and the sum of the first 12.

Solution

Each term is 3 times the one before (15 \div 5 = 45 \div 15 = 3), so the series is geometric with a_1 = 5 and r = 3. Since r > 1, the form with r^n - 1 keeps the numbers positive:

S_8 = \frac{5(3^8 - 1)}{3 - 1} = \frac{5 \cdot 6560}{2} = 16400

which agrees with the subtraction above. With n = 12 nothing changes but the exponent:

S_{12} = \frac{5(3^{12} - 1)}{2} = \frac{5 \cdot 531440}{2} = 1328600

Adding twelve terms by hand would take eleven additions; the formula takes the same work for 12 terms as for 8.

The sizes make sense. The 8th term alone is 5 \cdot 3^7 = 10935, and the seven terms before it add to 16400 - 10935 = 5465, about half the last term. When r = 3 the last term outweighs all the others together.

Example 2

A ratio less than 1

Find the sum of the first 7 terms of 96 + 48 + 24 + \cdots

Solution

Each term is half the one before, so a_1 = 96 and \displaystyle r = \frac{1}{2}. Since r < 1, the form with 1 - r keeps the brackets positive:

\begin{aligned} S_7 &= \frac{96\left(1 - \left(\frac{1}{2}\right)^7\right)}{1 - \frac{1}{2}} \\ &= \frac{96 \cdot \frac{127}{128}}{\frac{1}{2}} \end{aligned}

Dividing by \displaystyle \frac{1}{2} is multiplying by 2, so

S_7 = 192 \cdot \frac{127}{128} = \frac{381}{2} = 190.5

Check. The terms are 96, 48, 24, 12, 6, 3, 1.5, and they add to 190.5.

Notice that 190.5 = 192 - 1.5: the total is 192 less the last term. Each new term closes half of the gap still left below 192, so the totals creep up toward 192 without reaching it. What that means for infinitely many terms is the business of Infinite geometric series.

Example 3

A negative ratio in sigma notation

Find \sum_{k=1}^{10} 4(-2)^{k-1}.

Solution

The rule 4(-2)^{k-1} is a number times a power of -2, so the terms are geometric with r = -2. The first term is the value at k = 1, which is 4(-2)^0 = 4, and k runs from 1 to 10, so n = 10. The terms are 4, -8, 16, -32, \ldots, ending with 4(-2)^9 = -2048.

\begin{aligned} \sum_{k=1}^{10} 4(-2)^{k-1} &= \frac{4\left(1 - (-2)^{10}\right)}{1 - (-2)} \\ &= \frac{4(1 - 1024)}{3} = -1364 \end{aligned}

The brackets matter: (-2)^{10} = 1024 because the exponent is even, and 1 - (-2) = 3.

Check. The terms pair up as 4 - 8 = -4, 16 - 32 = -16, 64 - 128 = -64, 256 - 512 = -256 and 1024 - 2048 = -1024. These five pairs are themselves geometric with ratio 4, and -(4 + 16 + 64 + 256 + 1024) = -1364.

Example 4

Counting the terms first

Find 3 + 6 + 12 + \cdots + 1536.

Solution

The ratio is 2 and a_1 = 3. The last term is given but the number of terms is not. The term in position n is 3 \cdot 2^{n-1}, and it equals 1536 when 2^{n-1} = 512 = 2^9, so n - 1 = 9 and n = 10. Then

S_{10} = \frac{3(2^{10} - 1)}{2 - 1} = 3 \cdot 1023 = 3069

The last-term form gets there without n at all, since r\,a_n = 2 \cdot 1536 = 3072:

S = \frac{3 - 3072}{1 - 2} = \frac{-3069}{-1} = 3069

Example 5

Saving a fixed amount each year

Ezequiel pays 1200 dollars into a savings account at the end of every year. The account pays 4.6\% interest, compounded once a year. How much is in the account right after his 10th payment, and how much of it is interest?

Solution

Line the payments up by how long each has been growing. The 10th payment has just been made and is worth 1200. The 9th has earned one year of interest and is worth 1200(1.046). Each year the balance is multiplied by 1.046, and the 1st payment has been growing for 9 years, so it is worth 1200(1.046)^9. The balance is

1200 + 1200(1.046) + \cdots + 1200(1.046)^9

a geometric series with a_1 = 1200, r = 1.046 and n = 10 (exponents 0 to 9):

\begin{aligned} S_{10} &= \frac{1200(1.046^{10} - 1)}{1.046 - 1} \\ &\approx \frac{1200(0.567895)}{0.046} \approx 14814.64 \end{aligned}

There is about 14{,}814.64 dollars in the account. He paid in 10 \cdot 1200 = 12000 dollars, so about 2{,}814.64 dollars of it is interest. Keep 1.046^{10} unrounded on the calculator until the last step; rounding it early moves the answer by whole dollars.

Example 6

How many payments to reach a target

Ezequiel keeps the same plan. After which payment does the balance first reach 20{,}000 dollars?

Solution

After n payments the balance is \displaystyle \frac{1200(1.046^n - 1)}{0.046}, and the question is the smallest n that makes it at least 20000:

\frac{1200(1.046^n - 1)}{0.046} \ge 20000

Multiply both sides by the positive number \displaystyle \frac{0.046}{1200} and add 1:

\begin{aligned} 1.046^n &\ge 1 + \frac{20000 \cdot 0.046}{1200} \\ &= 1 + \frac{23}{30} = \frac{53}{30} \end{aligned}

Now n is in an exponent, so take a logarithm of both sides. The natural logarithm keeps the order of positive numbers, so \displaystyle 1.046^n \ge \frac{53}{30} becomes \displaystyle n \ln 1.046 \ge \ln \frac{53}{30}, and \ln 1.046 is positive, so dividing by it keeps the inequality the same way round:

n \ge \frac{\ln \frac{53}{30}}{\ln 1.046} \approx \frac{0.56909}{0.04497} \approx 12.65

(Common logarithms give the same quotient, \displaystyle \frac{0.24715}{0.01953}.)

Payments come in whole numbers, so the first that works is n = 13. Check either side: after 12 payments the balance is about 18{,}664.13 dollars, short of the target, and after 13 it is about 20{,}722.68 dollars. The balance first reaches 20{,}000 dollars with the 13th payment.

Example 7

The monthly payment on a loan

A loan of 8500 dollars is charged 6.6\% a year, compounded monthly, and is repaid in 36 equal monthly payments, the first one month after the loan is made. Find the payment.

Solution

Monthly compounding at 6.6\% a year means \displaystyle \frac{0.066}{12} = 0.0055 a month, a growth factor of 1.0055 each month.

Compare everything at the moment of the last payment, 36 months on. Had nothing been paid, the debt would have grown to 8500(1.0055)^{36}. Now think of each payment P as earning the same interest from the day it is paid until then. The last payment is worth P, the one before P(1.0055), and the first, 35 months earlier, P(1.0055)^{35}. The loan is cleared exactly when the payments, grown in this way, match the grown debt:

\begin{aligned} &P + P(1.0055) + \cdots + P(1.0055)^{35} \\ &\quad = 8500(1.0055)^{36} \end{aligned}

The left side is a geometric series with 36 terms, so

\frac{P(1.0055^{36} - 1)}{0.0055} = 8500(1.0055)^{36}

and solving for P,

\begin{aligned} P &= \frac{8500 \cdot 0.0055 \cdot 1.0055^{36}}{1.0055^{36} - 1} \\ &\approx \frac{56.9556}{0.218301} \approx 260.90 \end{aligned}

The payment is about 260.90 dollars a month. Over 36 months that is 36 \cdot 260.90 = 9392.40 dollars, so the loan costs about 892.40 dollars in interest.

Why compare at the end? Interest is charged on what is owed, and a payment reduces what is owed from the day it is made, so a payment made early saves exactly the interest it would have earned had it been saved instead. Moving every amount to the same date with the same factor 1.0055 makes them fair to compare.

Boundaries

The ratio must be the same all the way along. The cancellation in the subtraction works only because multiplying by r turns every term into the next one. For 1 + 2 + 4 + 7 the step factors are \displaystyle 2, 2, \frac{7}{4}, the series is not geometric, and the formula with r = 2 gives 15 against a true total of 14.

r = 1 is excluded. The formula divides by 1 - r. When r = 1 every term is a_1, and the sum is n a_1, which is also what the formula approaches for r close to 1. A first term of 0 makes every term 0 and the sum 0, which the formula gives correctly.

n counts terms. It must be a positive whole number. A sum written from k = 0 to k = 5 has 6 terms, and a list from a_1 to a given last term has n found from a_1 r^{n-1} = a_n. If that equation has no whole-number solution, the number given is not a term of the sequence.

A target may fall between two totals. Solving S_n \ge T for n usually gives a decimal, as in Example 6. The answer is then the next whole number up, and a check of the two totals either side confirms it. Dividing an inequality by \log r also flips it when 0 < r < 1, because then \log r is negative.

The money models assume what they say. The deposit and loan formulas assume equal amounts, paid at the end of each period, with the same growth factor every period and compounding once a period. A payment at the start of each period earns one extra period on every deposit and multiplies the total by g. A payment rounded to the cent leaves a few cents over at the end, which lenders settle in the final payment.

This is a finite sum. The formula adds the first n terms. Whether a geometric series of infinitely many terms has a total at all depends on the size of r, and that question belongs to Infinite geometric series.

Where this goes next

When -1 < r < 1 the power r^n in the formula shrinks toward 0 as n grows, and the totals settle down toward a single number, as the totals in Example 2 settled toward 192. That is the start of infinite geometric series, where a sum such as \displaystyle 12 + 4 + \frac{4}{3} + \cdots turns out to have a finite total, and where repeating decimals become fractions. Defining the sum of infinitely many terms as the limit of the partial sums is the subject of Infinite series and partial sums.

The deposit and loan calculations are the same series used again and again in personal finance: savings plans, mortgages and car loans all rest on \displaystyle \frac{P(g^n - 1)}{g - 1}. They join the single deposits of Compound interest and compounding periods to the repeated deposits of real accounts.

Exercises

Sums from the formula

Find the sum of the first 6 terms of 6, 24, 96, \ldots
Find the sum of the first 9 terms of 1, -3, 9, \ldots
Find the sum of the first 8 terms of 160, 80, 40, \ldots
Find the sum of the first 5 terms of \displaystyle 2, \frac{2}{3}, \frac{2}{9}, \ldots
Find the sum of the first 12 terms of 10, 10, 10, \ldots, and say why the sum formula cannot be used.

Sums in sigma notation

Find \sum_{k=1}^{7} 3 \cdot 2^{k}.
Find \displaystyle \sum_{k=0}^{5} 81 \left(\frac{1}{3}\right)^{k}.
Find \sum_{k=3}^{8} 2^{k}.

Counting the terms first

Find 4 + 12 + 36 + \cdots + 8748.
Find 1 - 2 + 4 - 8 + \cdots + 256.
Find 2048 + 1024 + 512 + \cdots + 1.

In a setting

A job starts at 43{,}500 dollars a year, and the pay rises by 2.5\% at the start of each later year. What is the total pay over the first 10 years?
At the end of every year 800 dollars is paid into an account that earns 3.4\% a year, compounded once a year. How much is in the account right after the 15th payment?
A ball is dropped from a height of 18 feet, and each rebound reaches \displaystyle \frac{2}{3} of the height of the fall before it. How far has the ball traveled, up and down, when it hits the ground for the 5th time?
One cent is put aside on day 1, two cents on day 2, four on day 3, and so on, doubling every day. How much has been put aside in total after 30 days? Give the answer in dollars.
A loan of 4800 dollars is charged 7.8\% a year, compounded monthly, and is repaid in 12 equal monthly payments, the first one month after the loan. Find the payment and the total interest.

Working backwards

A geometric series has 6 terms, common ratio 2 and sum 189. Find its first term.
What is the least number of terms of 6 + 12 + 24 + \cdots whose sum is more than 6000?
A geometric sequence has first term 2, and its first three terms add to 26. Find every possible common ratio, and write out the three terms for each.

Reasoning

Show that \displaystyle \frac{a_1(1 - r^n)}{1 - r} and \displaystyle \frac{a_1(r^n - 1)}{r - 1} are always equal, and explain what goes wrong with both when r = 1.
Starting from the sum formula, prove the last-term form \displaystyle S_n = \frac{a_1 - r\,a_n}{1 - r}.
Prove that 1 + 2 + 4 + \cdots + 2^{n-1} = 2^n - 1 twice: once with the sum formula, and once by explaining why adding the next term to both sides keeps the statement true, as Figure 1 suggests.
Suppose a_1 > 0 and 0 < r < 1. Show that every partial sum S_n is less than \displaystyle \frac{a_1}{1 - r}, and check this with the series of Example 2.

Answers

8190. a_1 = 6 and r = 4, so \displaystyle S_6 = \frac{6(4^6 - 1)}{4 - 1} = 2 \cdot 4095 = 8190.
4921. r = -3, so \displaystyle S_9 = \frac{1 - (-3)^9}{1 - (-3)}, which is \displaystyle \frac{1 + 19683}{4} = 4921.
\displaystyle \frac{1275}{4} = 318.75. \displaystyle r = \frac{1}{2}, so \displaystyle S_8 = \frac{160\left(1 - \frac{1}{256}\right)}{\frac{1}{2}}, which is \displaystyle 320 \cdot \frac{255}{256} = 318.75.
\displaystyle \frac{242}{81}. \displaystyle r = \frac{1}{3}, so \displaystyle S_5 = \frac{2\left(1 - \frac{1}{243}\right)}{\frac{2}{3}}, which is \displaystyle 3 \cdot \frac{242}{243} = \frac{242}{81}.
120. Every term is 10, so the sum is 12 \cdot 10 = 120. Here r = 1, and the formula would divide by 1 - r = 0.
762. The first term (k = 1) is 6, the ratio is 2 and there are 7 terms, so the sum is 6(2^7 - 1) = 6 \cdot 127 = 762.
\displaystyle \frac{364}{3}. The terms run from 81 (k = 0) to \displaystyle \frac{1}{3}, which is 6 terms with \displaystyle r = \frac{1}{3}: \displaystyle \frac{81\left(1 - \frac{1}{729}\right)}{\frac{2}{3}}, which is \displaystyle \frac{243}{2} \cdot \frac{728}{729} = \frac{364}{3}.
504. The terms are 8, 16, \ldots, 256, which is 8 - 3 + 1 = 6 terms with first term 8, so the sum is 8(2^6 - 1) = 504.
13120. r = 3 and 4 \cdot 3^{n-1} = 8748 gives 3^{n-1} = 2187 = 3^7, so n = 8; then \displaystyle S_8 = \frac{4(3^8 - 1)}{2} = 2 \cdot 6560 = 13120.
171. r = -2 and (-2)^{n-1} = 256 gives n - 1 = 8, so n = 9; then \displaystyle S_9 = \frac{1 - (-2)^9}{1 - (-2)} = \frac{513}{3} = 171.
4095. \displaystyle r = \frac{1}{2} and 2048 = 2^{11}, so the terms run from 2^{11} down to 2^0, which is n = 12; then \displaystyle S_{12} = \frac{2048\left(1 - \frac{1}{4096}\right)}{\frac{1}{2}} = 4095.
About 487{,}347.11 dollars. The yearly pay is geometric with a_1 = 43500 and r = 1.025, so the total is \displaystyle \frac{43500(1.025^{10} - 1)}{0.025} \approx 487347.11.
About 15{,}323.10 dollars. The payments are worth 800, 800(1.034), \ldots, 800(1.034)^{14}, so the balance is \displaystyle \frac{800(1.034^{15} - 1)}{0.034} \approx 15323.10.
\displaystyle \frac{682}{9} \approx 75.78 feet. The ball falls 18, then rises and falls each of \displaystyle 12, 8, \frac{16}{3}, \frac{32}{9} before the 5th landing. Those four heights add to \displaystyle \frac{12\left(1 - \left(\frac{2}{3}\right)^4\right)}{\frac{1}{3}} = \frac{260}{9}, so the distance is \displaystyle 18 + 2 \cdot \frac{260}{9} = \frac{682}{9}.
10{,}737{,}418.23 dollars. The total in cents is 1 + 2 + \cdots + 2^{29} = 2^{30} - 1 = 1073741823.
About 417.10 dollars a month, and about 205.20 dollars in interest. The monthly factor is \displaystyle 1 + \frac{0.078}{12} = 1.0065, and as in Example 7 the payment P is \displaystyle \frac{4800 \cdot 0.0065 \cdot 1.0065^{12}}{1.0065^{12} - 1}, about 417.10; then 12 \cdot 417.10 = 5005.20, which is 205.20 more than 4800.
3. \displaystyle S_6 = \frac{a_1(2^6 - 1)}{2 - 1} = 63a_1 = 189, so a_1 = 3.
10 terms. S_n = 6(2^n - 1) > 6000 needs 2^n > 1001. Since 2^9 = 512 and 2^{10} = 1024, the least is n = 10: S_9 = 3066 and S_{10} = 6138.
r = 3 or r = -4. 2 + 2r + 2r^2 = 26 gives r^2 + r - 12 = 0, which is (r + 4)(r - 3) = 0. The terms are 2, 6, 18 or 2, -8, 32, and both add to 26.
Multiply the top and bottom of the first by -1: the top a_1(1 - r^n) becomes a_1(r^n - 1) and the bottom 1 - r becomes r - 1, and multiplying top and bottom by the same nonzero number leaves a fraction unchanged. When r = 1 both tops and both bottoms are 0, so neither expression has a value. The subtraction step that led to them, S_n(1 - r) = a_1(1 - r^n), reads 0 = 0 and says nothing about S_n; the sum is found directly as n a_1.
The last term is a_n = a_1 r^{n-1}, so r\,a_n = a_1 r^n. Then a_1(1 - r^n) expands to a_1 - a_1 r^n, which is a_1 - r\,a_n, and dividing by 1 - r gives the last-term form.
By the formula with a_1 = 1 and r = 2: \displaystyle \frac{2^n - 1}{2 - 1} = 2^n - 1. Second way: the statement holds for n = 1, since 1 = 2^1 - 1. If 1 + \cdots + 2^{n-1} = 2^n - 1, adding the next term 2^n to both sides gives 2^n - 1 + 2^n on the right, which is 2 \cdot 2^n - 1, or 2^{n+1} - 1: the same statement one term further on. So every stack stays exactly one short of the next block, and the statement holds for every n.
Split the formula into two fractions:

S_n = \frac{a_1}{1 - r} - \frac{a_1 r^n}{1 - r}

With a_1 > 0 and 0 < r < 1, the subtracted part has a positive top and a positive bottom, so it is positive and \displaystyle S_n < \frac{a_1}{1 - r}. In Example 2, \displaystyle \frac{96}{1 - \frac{1}{2}} = 192, and indeed S_7 = 190.5 < 192; the amount short is \displaystyle \frac{96 \cdot \left(\frac{1}{2}\right)^7}{\frac{1}{2}} = 1.5.