10 Sequences and Series

A structured guide to identifying, representing, modeling, and summing arithmetic and geometric sequences and series.

Core Ideas: Sequences and

A is an ordered list of numbers, while a is formed by adding terms. The terms are labeled by position: the term in position nn is written as ana_n. For example, the list 2,5,8,11,…2,5,8,11,\ldots has a1=2a_1=2, a2=5a_2=5, and a3=8a_3=8.

A can be described in several ways:

  • A list of terms shows the pattern directly.

  • A table or graph connects positions with values.

  • A explains how to generate each term from earlier terms.

  • An gives a term directly from its position.

A may be written in expanded form, such as a1+a2+a3a_1+a_2+a_3, or compactly with .

Takeaway: A describes values in order; a describes their sum.

Two Ways to Describe a

A recursive description requires an initial value and a rule for producing later terms. For the 4,7,10,13,…4,7,10,13,\ldots, one recursive representation is

a1=4,an=an−1+3(n≥2).a_1=4,\qquad a_n=a_{n-1}+3\quad(n\ge2).

The terms are generated in order:

a2=4+3=7,a3=7+3=10,a4=10+3=13.a_2=4+3=7,\qquad a_3=7+3=10,\qquad a_4=10+3=13.

An explicit description gives the same directly:

an=3n+1.a_n=3n+1.

Thus, a distant term can be found without calculating all the previous terms:

a20=3(20)+1=61.a_{20}=3(20)+1=61.

Use a when the generation process matters or when terms are built sequentially. Use an when a particular, especially distant, term is needed efficiently.

Takeaway: Recursive formulas emphasize construction; explicit formulas emphasize direct calculation.

Arithmetic Patterns

An has a constant additive change between consecutive terms. This constant is the , denoted by dd:

d=an−an−1.d=a_n-a_{n-1}.

For example, 12,17,22,27,…12,17,22,27,\ldots is arithmetic because each term increases by 55. Its recursive form is

a1=12,an=an−1+5.a_1=12,\qquad a_n=a_{n-1}+5.

Its explicit form is

an=a1+(n−1)d.a_n=a_1+(n-1)d.

The factor n−1n-1 counts the changes made after the first term. For 7,11,15,19,…7,11,15,19,\ldots, the first term is 77 and the is 44, so

a25=7+(25−1)(4)=103.a_{25}=7+(25-1)(4)=103.

If two terms are known, the can be calculated using

d=aj−aij−i.d=\frac{a_j-a_i}{j-i}.

Takeaway: Constant subtraction of consecutive terms identifies an , and its terms follow an=a1+(n−1)da_n=a_1+(n-1)d.

Geometric Patterns

A has a constant multiplicative change between consecutive terms. This factor is the , denoted by rr:

r=anan−1.r=\frac{a_n}{a_{n-1}}.

For example, 3,12,48,192,…3,12,48,192,\ldots is geometric because each term is multiplied by 44. Its recursive form is

a1=3,an=4an−1,a_1=3,\qquad a_n=4a_{n-1},

and its explicit form is

an=a1rn−1.a_n=a_1r^{n-1}.

For 5,10,20,40,…5,10,20,40,\ldots, the first term is 55 and the is 22. Therefore,

a6=5(2)6−1=160.a_6=5(2)^{6-1}=160.

The value of the ratio determines the pattern:

  • If ∣r∣>1\lvert r\rvert>1, the magnitudes generally grow.

  • If 0<∣r∣<10<\lvert r\rvert<1, the magnitudes generally decay.

  • If r<0r<0, the signs alternate.

Takeaway: Constant division of consecutive terms identifies a , and its terms follow an=a1rn−1a_n=a_1r^{n-1}.

Compact Addition with Sigma Notation

uses the Greek letter sigma to express a long addition compactly. In

∑k=1nak,\sum_{k=1}^{n}a_k,

kk is the index of summation, 11 is the starting value, nn is the ending value, and aka_k is the general term.

For example,

∑k=15(2k+1)=(2⋅1+1)+(2⋅2+1)+⋯+(2⋅5+1).\sum_{k=1}^{5}(2k+1)=(2\cdot1+1)+(2\cdot2+1)+\cdots+(2\cdot5+1).

Evaluating the terms gives

3+5+7+9+11=35.3+5+7+9+11=35.

An arithmetic list can also be represented with :

4+7+10+13+16=∑k=15(4+3(k−1)).4+7+10+13+16=\sum_{k=1}^{5}\bigl(4+3(k-1)\bigr).

Takeaway: Read a sigma expression by identifying its index, starting value, ending value, and general term.

Adding Finite Patterns

An adds the first nn terms of an . Its main sum formula is

Sn=n2(a1+an).S_n=\frac{n}{2}(a_1+a_n).

When the is known, use the equivalent form

Sn=n2[2a1+(n−1)d].S_n=\frac{n}{2}\left[2a_1+(n-1)d\right].

These formulas work because terms can be paired from opposite ends, and each pair has the same sum.

For the 6,10,14,18,…6,10,14,18,\ldots, the first term is 66, the is 44, and the twentieth term is

a20=6+(20−1)(4)=82.a_{20}=6+(20-1)(4)=82.

Therefore, the sum of the first twenty terms is

S20=202(6+82)=880.S_{20}=\frac{20}{2}(6+82)=880.

A adds a fixed number of terms of a . For r≠1r\ne1, use

Sn=a11−rn1−r.S_n=a_1\frac{1-r^n}{1-r}.

For the first eight terms of 2,6,18,54,…2,6,18,54,\ldots, where a1=2a_1=2 and r=3r=3,

S8=21−381−3=6560.S_8=2\frac{1-3^8}{1-3}=6560.

If r=1r=1, every term equals a1a_1, so Sn=na1S_n=na_1.

Takeaway: Use the arithmetic sum formula for constant differences and the finite geometric sum formula for constant ratios.

Infinite Geometric Sums

An has a finite sum only when the satisfies

∣r∣<1.\lvert r\rvert<1.

In that case,

S∞=a11−r.S_\infty=\frac{a_1}{1-r}.

For example, the

12+6+3+32+⋯12+6+3+\frac{3}{2}+\cdots

has a1=12a_1=12 and r=12r=\frac{1}{2}, so

S∞=121−12=24.S_\infty=\frac{12}{1-\frac{1}{2}}=24.

When ∣r∣≥1\lvert r\rvert\ge1, the terms do not approach zero in the required way, so the infinite does not converge to a finite sum.

Takeaway: Before applying the infinite-sum formula, always check that ∣r∣<1\lvert r\rvert<1.

Selecting and Applying a Model

Choose a model by examining how the quantity changes from one step to the next.

  1. Subtract consecutive values. A constant difference suggests an arithmetic model.

  2. Divide consecutive nonzero values. A constant ratio suggests a geometric model.

  3. Check the context. Fixed additions or subtractions suggest arithmetic change, while repeated percentage changes suggest geometric change.

  4. Use the selected formula to calculate or predict values, remembering that real data may only approximately follow a pattern.

For a constant-addition situation, such as weekly deposits beginning at 4040 dollars and increasing by 55 dollars each week,

an=40+(n−1)(5).a_n=40+(n-1)(5).

The twelfth deposit is

a12=40+11(5)=95,a_{12}=40+11(5)=95,

and the total deposited over twelve weeks is

S12=122(40+95)=810.S_{12}=\frac{12}{2}(40+95)=810.

For a repeated percentage decrease, convert the percentage to a multiplier. A decrease of 1212 percent means multiplying by 0.880.88. If the initial value is 18,00018{,}000, the value after nn years is modeled by

Vn=18,000(0.88)n.V_n=18{,}000(0.88)^n.

Takeaway: Constant additive change calls for an arithmetic model; constant multiplicative or percentage change calls for a geometric model.