Unit 9, lesson 105

Geometric Sequences and Series

How a constant ratio gives a formula for any term of a geometric sequence, and how multiplying a sum by that ratio and subtracting adds up any number of its terms in one step.

  • Theory part 209
  • Practice part 210
  • Series

Fold a sheet of paper in half and it is 2 layers thick. Fold it again and it is 4 layers, then 8, then 16. Each number is the one before it times 2. Lists that grow or shrink by the same factor at every step turn up in interest, population growth and the way medicine leaves the body. A short formula gives any term of such a list, and one subtraction adds up as many terms as you want.

Key terms

Common ratio

Definition A sequence {an}\{a_n\} is geometric if its first term is not zero and there is a real number r≠0r \ne 0, called the common ratio, such that

an+1=r ana_{n+1} = r\,a_n

for every index nn for which n+1n + 1 is also an index. Then every term is nonzero, and the ratio of each term to the one before it is

an+1an=r.\frac{a_{n+1}}{a_n} = r.

Some books, and the calculus lessons, also allow r=0r = 0, which gives the list a1,0,0,…a_1, 0, 0, \dots. The precalculus lessons leave it out, so dividing by a term is always allowed.

In plain words Every step multiplies by the same number rr. When ∣r∣<1\lvert r \rvert < 1 the terms shrink in size, when ∣r∣>1\lvert r \rvert > 1 they grow in size, and a negative rr makes the signs alternate.

Example The list 5,15,45,1355, 15, 45, 135 has the ratios 15/515/5, 45/1545/15 and 135/45135/45, all equal to 33, so it is geometric with r=3r = 3. The list 48,−24,12,−648, -24, 12, -6 is geometric with r=−12r = -\frac{1}{2}. The list 2,4,6,82, 4, 6, 8 is arithmetic but not geometric, because 4/2=24/2 = 2 while 6/4=1.56/4 = 1.5.

Explicit nth term

Definition The geometric sequence with first term a1≠0a_1 \ne 0 and common ratio r≠0r \ne 0 has the explicit formula

an=a1rn−1for n≥1.a_n = a_1 r^{n-1} \quad \text{for } n \ge 1.

If the counting starts at n=0n = 0 with a starting value a0a_0, the formula is an=a0rna_n = a_0 r^n. For any two indices mm and nn,

an=amrn−m.a_n = a_m r^{n-m}.

When r>0r > 0 and r≠1r \ne 1, the formula can be written an=a1r⋅rna_n = \frac{a_1}{r} \cdot r^n, so the points (n,an)(n, a_n) lie on the graph of the exponential function y=a1r⋅rxy = \frac{a_1}{r} \cdot r^x. A geometric sequence is the whole-number version of exponential growth and decay. The common ratio plays the role of the growth factor bb there, not of the growth rate, so growth of 5 percent per step means r=1.05r = 1.05.

In plain words To get from the first term to the nnth term you take n−1n - 1 steps, and each step multiplies by rr. So the exponent is n−1n - 1, not nn.

Example For 5,15,45,…5, 15, 45, \dots the formula is an=5⋅3n−1a_n = 5 \cdot 3^{n-1}, so a8=5⋅37=10935a_8 = 5 \cdot 3^7 = 10935 without listing the terms in between. Two terms are also enough to find the sequence. If a2=6a_2 = 6 and a5=162a_5 = 162, the three steps from a2a_2 to a5a_5 give

r3=1626=27,r^3 = \frac{162}{6} = 27,

so r=3r = 3 and a1=6/3=2a_1 = 6/3 = 2. With an even number of steps, watch the sign. An equation like r2=9r^2 = 9 allows both r=3r = 3 and r=−3r = -3.

Finite geometric sum

Definition The sum of the first nn terms of the geometric sequence with first term a1a_1 and common ratio rr is

Sn=a1+a1r+a1r2+⋯+a1rn−1=∑k=1na1rk−1.\begin{aligned} S_n &= a_1 + a_1 r + a_1 r^2 \\ &\quad + \cdots + a_1 r^{n-1} \\[0.8em] &= \sum_{k=1}^{n} a_1 r^{k-1}. \end{aligned}

If r≠1r \ne 1, then

Sn=a1(1−rn)1−r=a1(rn−1)r−1.S_n = \frac{a_1(1 - r^n)}{1 - r} = \frac{a_1(r^n - 1)}{r - 1}.

The second form is the first with top and bottom multiplied by −1-1, and it avoids negative numbers when r>1r > 1. The next section shows where the formula comes from.

In plain words Multiply the first term by 1−rn1 - r^n and divide by 1−r1 - r, where nn is the number of terms. That exponent is one more than the exponent on the last term, because the last term is a1rn−1a_1 r^{n-1}.

Example The sum 2+6+18+⋯+43742 + 6 + 18 + \cdots + 4374 has a1=2a_1 = 2 and r=3r = 3. Its last term is 4374=2⋅374374 = 2 \cdot 3^7, so there are 88 terms and

S8=2(38−1)3−1=38−1=6560.S_8 = \frac{2(3^8 - 1)}{3 - 1} = 3^8 - 1 = 6560.

A shrinking list works the same way. The sum 64+32+16+⋯+164 + 32 + 16 + \cdots + 1 has a1=64a_1 = 64, r=12r = \frac{1}{2} and 77 terms, so

S7=64(1−(12)7)1−12=128(1−1128)=127.\begin{aligned} S_7 &= \frac{64\left(1 - \left(\frac{1}{2}\right)^7\right)}{1 - \frac{1}{2}} \\[0.8em] &= 128\left(1 - \frac{1}{128}\right) \\[0.8em] &= 127. \end{aligned}

Derivation by subtraction

Definition For every real number rr and every integer n≥1n \ge 1, the sum Sn=a1+a1r+⋯+a1rn−1S_n = a_1 + a_1 r + \cdots + a_1 r^{n-1} satisfies

(1−r) Sn=a1(1−rn).(1 - r)\,S_n = a_1(1 - r^n).

To see why, multiply the sum by rr and write it under the original,

Sn=a1+a1r+a1r2+⋯+a1rn−1,rSn=a1r+a1r2+⋯+a1rn−1+a1rn.\begin{aligned} S_n &= a_1 + a_1 r + a_1 r^2 \\ &\quad + \cdots + a_1 r^{n-1}, \\[0.4em] r S_n &= a_1 r + a_1 r^2 \\ &\quad + \cdots + a_1 r^{n-1} + a_1 r^n. \end{aligned}

Subtracting the second line from the first cancels every term except a1a_1 and −a1rn-a_1 r^n, so Sn−rSn=a1−a1rnS_n - rS_n = a_1 - a_1 r^n. When r≠1r \ne 1, dividing by 1−r1 - r gives the sum formula. With a1=1a_1 = 1 the same identity is a factorization that holds for every rr,

1−rn=(1−r)(1+r+r2+⋯+rn−1).\begin{aligned} 1 - r^n &= (1 - r)(1 + r + r^2 \\ &\qquad + \cdots + r^{n-1}). \end{aligned}

In plain words Multiplying by rr shifts every term one place along. The sum and its shifted copy share all their middle terms, so subtracting leaves only the first term of one and the last term of the other.

Example Take S=1+3+9+27+81S = 1 + 3 + 9 + 27 + 81 and multiply by the ratio 33,

3S=3+9+27+81+243,S=1+3+9+27+81.\begin{aligned} 3S &= 3 + 9 + 27 + 81 + 243, \\ S &= 1 + 3 + 9 + 27 + 81. \end{aligned}

Subtracting the second line from the first cancels the four shared terms, so 2S=243−1=2422S = 243 - 1 = 242 and S=121S = 121. Adding the five numbers directly gives 121121 as well.

Ratio equal to 1

Definition If r=1r = 1, every term equals a1a_1 and

Sn=a1+a1+⋯+a1=n a1.S_n = a_1 + a_1 + \cdots + a_1 = n\,a_1.

The formula with 1−r1 - r in the denominator does not apply, because at r=1r = 1 it becomes 00\frac{0}{0}, which is undefined. The identity (1−r)Sn=a1(1−rn)(1 - r)S_n = a_1(1 - r^n) still holds, but both sides are 00, so it says nothing about SnS_n. No other ratio needs special care. For r=−1r = -1, for example, the formula gives

Sn=a1(1−(−1)n)2,S_n = \frac{a_1\left(1 - (-1)^n\right)}{2},

which is a1a_1 for odd nn and 00 for even nn, exactly what direct adding shows.

In plain words With ratio 11 nothing changes from term to term, so the sum is the number of terms times the first term. The general formula would divide by zero there, so this one ratio gets its own rule.

Example The sum 7+7+7+7+77 + 7 + 7 + 7 + 7 is 5⋅7=355 \cdot 7 = 35. A ratio close to 11 gives a sum close to this. With a1=7a_1 = 7, r=1.01r = 1.01 and n=5n = 5 the formula gives

7(1.015−1)1.01−1≈35.71.\frac{7(1.01^5 - 1)}{1.01 - 1} \approx 35.71.

So the formula and the special case fit together, and only the exact value r=1r = 1 has to be handled separately.

Applications

Definition A quantity that is multiplied by the same factor rr in each period forms a geometric sequence, and a total collected over several periods is a finite geometric sum. A standard case is a savings plan. Suppose a deposit PP is made at the end of each period into an account that earns interest at the rate i>0i > 0 per period, compounded once per period. Right after the nnth deposit the newest deposit has earned nothing and the first has grown for n−1n - 1 periods, so the balance is

Bn=P+P(1+i)+⋯+P(1+i)n−1=P⋅(1+i)n−1i.\begin{aligned} B_n &= P + P(1 + i) + \cdots \\ &\quad + P(1 + i)^{n-1} \\[0.8em] &= P \cdot \frac{(1 + i)^n - 1}{i}. \end{aligned}

In plain words Look for a fixed growth or shrink factor. The amounts at each step form a geometric sequence, and the total over several steps is a geometric sum with that factor as its ratio.

Example Someone puts 1000 dollars into an account at the end of every year, and the account earns 4 percent a year. The ratio is 1.041.04, and right after the tenth deposit the balance is

B10=1000⋅1.0410−10.04≈12006.11 dollars.\begin{aligned} B_{10} &= 1000 \cdot \frac{1.04^{10} - 1}{0.04} \\[0.8em] &\approx 12006.11 \text{ dollars.} \end{aligned}

Of that, 10000 dollars are deposits and about 2006 dollars are interest.

Check yourself

  1. A geometric sequence has a2=5a_2 = 5 and a4=45a_4 = 45. Find every possible value of rr, a1a_1 and a5a_5.
Show the answer

Two steps lead from a2a_2 to a4a_4, so r2=45/5=9r^2 = 45/5 = 9 and r=3r = 3 or r=−3r = -3. With r=3r = 3 the sequence has a1=53a_1 = \frac{5}{3} and a5=135a_5 = 135. With r=−3r = -3 it has a1=−53a_1 = -\frac{5}{3} and a5=−135a_5 = -135. Both sequences fit the given terms.

  1. Evaluate the sum below.
∑k=395⋅2k\sum_{k=3}^{9} 5 \cdot 2^k
Show the answer

The terms run from 4040 to 25602560 with ratio 22, and there are 9−3+1=79 - 3 + 1 = 7 of them. The sum is

40(27−1)2−1=40⋅127=5080.\frac{40(2^7 - 1)}{2 - 1} = 40 \cdot 127 = 5080.
  1. A job pays 40000 dollars in the first year, and the pay rises by 3 percent every year. What is the total pay over the first 10 years?
Show the answer

The yearly pay forms a geometric sequence with a1=40000a_1 = 40000 and r=1.03r = 1.03, so the total is

40000(1.0310−1)1.03−1≈458555.17 dollars.\begin{aligned} &\frac{40000(1.03^{10} - 1)}{1.03 - 1} \\[0.8em] &\quad \approx 458555.17 \text{ dollars.} \end{aligned}

How this lesson connects

Where this leads in Calculus