Contents / Precalculus / Sequences, Series & Conics

Chapter 8

Sequences, Series & Conics

Sequences, arithmetic and geometric progressions, the binomial theorem, and conic sections.

Introduction

Sequences, arithmetic and geometric progressions, the binomial theorem, and conic sections.

8.1Sequence

A sequence is an ordered list indexed by the natural numbers. It may converge to a limit or diverge.

Sequences

Definition 8.1 (Sequence). A sequence is a function whose domain is the positive integers (or nonnegative integers), written as an ordered list $a_{1}, a_{2}, a_{3}, \dots$ . Its $n$ th term is denoted $a_{n}$ .

Sequences may be defined explicitly (closed form) or recursively ( $a_{n + 1}$ in terms of previous terms). Intuitively, a sequence converges to a number $L$ if its terms eventually get and stay as close to $L$ as you like. Made precise:

Definition 8.2 (Convergence of a sequence). A sequence converges to $L$ if for every $ε > 0$ there exists $N$ such that

n > N ⟹ ∣ a_{n} - L ∣ < ε .

Otherwise the sequence diverges.

Here $ε$ is the "how close" tolerance you demand, and $N$ is how far out you must go before the tail stays within that tolerance.

Intuition. A sequence is just a list of numbers indexed by 1,2,3,… like the daily temperature readings forever. Convergence is like a dart that keeps landing closer and closer to the bullseye: no matter how tiny a circle you draw around the target, eventually every dart lands inside it. We care about whether the numbers settle down to a single value as we go further out in the list.

Example 8.3 (The sequence defined by $a_{n} = \frac{n}{n + 1}$ . Does it converge? To what?). Write the first terms: $\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots$ — they are climbing, but each stays below $1$ .

Solution.

Rewrite in a form where the behavior is obvious: $a_{n} = \frac{n + 1 - 1}{n + 1} = 1 - \frac{1}{n + 1}$ .
As $n \to \infty$ , the leftover $\frac{1}{n + 1} \to 0$ , so $a_{n} \to 1$ .
Therefore $lim_{n \to \infty} a_{n} = 1$ ; the sequence converges to $1$ .
Sanity check: $a_{99} = \frac{99}{100} = 0.99$ , already within $0.01$ of $1$ , and every later term is even closer. ✓

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Common Mistakes

Sequence vs series: A sequence is the list; a series is the sum of the list. Confusing $a_{n}$ with $S_{n}$ is common.
Convergence requires eventual closeness: Early terms can be wild; only the tail matters.
Bounded does not imply convergent: (-1)^n is bounded but diverges.
Monotonic + bounded implies convergent: Important theorem; students forget one or the other.
Limit of sequence vs limit of function: Similar but n integer; use same tools often (e.g. 1/n →0).
Recursive without base case: Need initial term(s) to generate the list.

Exercises for §8.1 →

8.2Arithmetic progression

Constant difference between terms. Simple closed forms for nth term and sum of first n terms.

Arithmetic Progressions

Definition 8.4 (Arithmetic progression). An arithmetic progression (AP) has a constant common difference $d$ between consecutive terms. To reach the $n$ th term you start at $a_{1}$ and add $d$ a total of $n - 1$ times, giving the closed form

a_{n} = a_{1} + (n - 1) d .

Proposition 8.5 (Sum of an arithmetic progression). The sum of the first $n$ terms, written $S_{n} = a_{1} + a_{2} + \dots + a_{n}$ , is

S_{n} = \frac{n}{2} (2 a_{1} + (n - 1) d) or equivalently S_{n} = \frac{n}{2} (first + last) .

Proof. The two forms agree because the last term is $a_{n} = a_{1} + (n - 1) d$ , so $first + last = a_{1} + a_{1} + (n - 1) d = 2 a_{1} + (n - 1) d$ . For the formula itself, pair first $+$ last, second $+$ second-last, and so on. Each such pair sums to the same value $2 a_{1} + (n - 1) d$ , and there are $\frac{n}{2}$ pairs.∎

Intuition. An arithmetic sequence is "add the same amount every step" — like saving a fixed $50 per month, so your balance is $50, $100, $150, … The sum formula is the trick Gauss used as a schoolboy: to add $1 + 2 + \dots + 100$ , fold the list so $1$ pairs with $100$ , $2$ with $99$ , …, each pair making $101$ ; fifty pairs give $50 \times 101 = 5050$ without adding one by one.

Example 8.6 (Find the sum of the first 15 terms of $4, 9, 14, \dots$ ). Identify $a_{1} = 4$ and $d = 9 - 4 = 5$ (check: $14 - 9 = 5$ ✓).

Solution.

Find the last term: $a_{15} = a_{1} + (15 - 1) d = 4 + 14 \cdot 5 = 4 + 70 = 74$ .
Apply the sum formula: $S_{15} = \frac{15}{2} (first + last) = \frac{15}{2} (4 + 74) = \frac{15}{2} \cdot 78 = 15 \cdot 39 = 585$ .
Sanity check with the other form: $S_{15} = \frac{15}{2} (2 \cdot 4 + 14 \cdot 5) = \frac{15}{2} (8 + 70) = \frac{15}{2} \cdot 78 = 585$ . ✓

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Common Mistakes

Arithmetic vs geometric sum formulas: Using n/2 (first+last) for a geometric series or a1(1-r^n)/(1-r) for arithmetic produces wrong answers.
n starts at 1: $a_{n} = a_{1} + (n - 1) d$ ; off-by-one common.
Sum to n or to infinity: AP infinite sum diverges unless d=0.
Common difference sign: d can be negative (decreasing AP).
Forgetting the /2 in sum: n/2 * (first+last).
Using last term without computing: last = a1 + (n-1)d ; don't guess.

Exercises for §8.2 →

8.3Geometric progression

Constant ratio between terms. Finite and infinite sum formulas (when |r|<1).

Geometric Progressions

Definition 8.7 (Geometric progression). A geometric progression (GP) has a constant common ratio $r$ : each term is the previous one multiplied by $r$ , so

a_{n} = a_{1} \cdot r^{n - 1} .

Proposition 8.8 (Sum of a finite geometric progression). For $r \neq = 1$ , the sum of the first $n$ terms is

S_{n} = a_{1} \frac{r ^{n} - 1}{r - 1} = a_{1} \frac{1 - r ^{n}}{1 - r} .

(The two forms are the same fraction with numerator and denominator both negated.)

Proof. This comes from a telescoping trick: writing $S_{n}$ and $r S_{n}$ and subtracting cancels every middle term, leaving $S_{n} - r S_{n} = a_{1} - a_{1} r^{n}$ , then divide by $1 - r$ .∎

Theorem 8.9 (Infinite geometric series). When $∣ r ∣ < 1$ the infinite geometric series converges, and

S_{\infty} = \frac{a _{1}}{1 - r} .

Proof. The condition $∣ r ∣ < 1$ is what makes it converge: then $r^{n} \to 0$ as $n \to \infty$ , so the finite sum $S_{n} = a_{1} \frac{1 - r ^{n}}{1 - r}$ tends to $\frac{a _{1}}{1 - r}$ .∎

When $∣ r ∣ \geq 1$ the terms do not shrink, the partial sums grow without bound, and the infinite sum diverges (the only exception being the trivial $r = 1$ case, where all terms equal $a_{1}$ and the finite sums are just $n a_{1}$ ).

Intuition. Geometric is "multiply by the same factor every step" — compound interest, half-life decay, or a bouncing ball that keeps half its height each bounce. The infinite sum formula feels like magic: even though you add forever, the total stays finite when $∣ r ∣ < 1$ . Think of walking half the remaining distance to a wall, then half of what's left, then half again — you add infinitely many steps but never travel more than the full distance to the wall.

Example 8.10 (Infinite sum of $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \dots$ ). Identify $a_{1} = \frac{1}{2}$ and $r = \frac{1/4}{1/2} = \frac{1}{2}$ .

Solution.

Check convergence: $∣ r ∣ = \frac{1}{2} < 1$ , so the infinite sum formula applies.
Apply it: $S_{\infty} = \frac{a _{1}}{1 - r} = \frac{1/2}{1 - 1/2} = \frac{1/2}{1/2} = 1$ .
Sanity check with partial sums: $\frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \dots$ climb toward $1$ but never reach it. You can also read it as the binary fraction $0.1111 \dots_{2} = 1$ . ✓

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Common Mistakes

Arithmetic vs geometric sum formulas: Using n/2 (first+last) for a geometric series or a1(1-r^n)/(1-r) for arithmetic produces wrong answers.
Infinite geometric only when |r|<1: When |r|≥1 the partial sums grow without bound; the formula a1/(1-r) does not apply.
r=1 special case: All terms equal a1; sum = n a1 (finite) or diverges (infinite).
Sign of r: |r|<1 for convergence; negative r alternates but still converges if |r|<1.
First term a1 or a0: Formula uses a1 the first term.
Confusing $S_{n}$ with $a_{n}$ : $S_{n}$ is the sum up to $n$ ; $a_{n}$ the $n$ th term.

Exercises for §8.3 →

8.4Binomial theorem

The binomial theorem gives a direct formula for expanding (x+y)^n without multiplying n times, using binomial coefficients.

Binomial Theorem

Theorem 8.11 (Binomial theorem). For a nonnegative integer $n$ ,

(x + y)^{n} = k = 0 \sum n (k n) x^{n - k} y^{k},

a sum of $n + 1$ terms, where the binomial coefficient $(k n) = \frac{n !}{k ! ( n - k )!}$ (read " $n$ choose $k$ ") counts the number of ways to pick $k$ items from $n$ .

Proof. Write $(x + y)^{n}$ as $n$ identical factors $(x + y) (x + y) \dots (x + y)$ . A term with $y^{k}$ is formed by choosing the $y$ from exactly $k$ of the factors (and $x$ from the other $n - k$ ); there are $(k n)$ such choices, so that is the coefficient. This combinatorial counting is one proof; the other standard proof is induction on $n$ .∎

In each term the exponents of $x$ and $y$ always add up to $n$ . Pascal's triangle gives a quick way to read the coefficients without computing factorials — each entry is the sum of the two above it, since $(k n) = (k - 1 n - 1) + (k n - 1)$ .

Intuition. Instead of multiplying $(x + y)$ by itself $n$ times, the theorem hands you each term's coefficient directly. Picture $n$ coins in a row; each coin you can flip to " $y$ " or leave as " $x$ ". A term like $x^{n - k} y^{k}$ is "exactly $k$ coins landed on $y$ ", and the coefficient $(k n)$ is just how many ways that can happen.

Example 8.12 (Expand $(x + 2)^{4}$ and identify the $x^{2}$ term.). Here $y = 2$ and $n = 4$ , so the term for each $k$ is $(k 4) x^{4 - k} 2^{k}$ :

Solution. $k = 0 : (0 4) x^{4} \cdot 2^{0} = 1 \cdot x^{4} \cdot 1 = x^{4}$ . $k = 1 : (1 4) x^{3} \cdot 2^{1} = 4 x^{3} \cdot 2 = 8 x^{3}$ . $k = 2 : (2 4) x^{2} \cdot 2^{2} = 6 x^{2} \cdot 4 = 24 x^{2}$ . $k = 3 : (3 4) x^{1} \cdot 2^{3} = 4 x \cdot 8 = 32 x$ . $k = 4 : (4 4) x^{0} \cdot 2^{4} = 1 \cdot 16 = 16$ . Full expansion: $x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16$ ; the $x^{2}$ coefficient is $24$ . Sanity check: the coefficients $1, 4, 6, 4, 1$ are exactly row $n = 4$ of Pascal's triangle. ✓□

Common Mistakes

Binomial coefficients: C(n,k)=C(n,n-k); forgetting this or miscalculating factorials for larger n.
Off by one in k: The term with y^k is C(n,k) x^{n-k} y^k ; power on x is n-k.
Forgetting the y^k power: Even if y=1, the coeff is still C(n,k).
Negative or fractional n: The basic theorem is for nonnegative integer n; generalized binomial for others.
Not using Pascal for small n: Quick to build row by row instead of computing large factorials.
Sign errors with (x - y): Alternating signs from y^k with negative.

Exercises for §8.4 →

8.5Conic section

Conics unify circles, ellipses, parabolas, hyperbolas as quadratic curves or cone slices. Standard forms after completing square/rotation.

Conic Sections

Definition 8.13 (Conic section). Conic sections are the curves obtained by slicing a double cone with a plane at various angles:

Circle: plane perpendicular to cone axis.
Ellipse: plane tilted but cuts only one nappe completely.
Parabola: plane parallel to a generator line.
Hyperbola: plane cuts both nappes.

Algebraically they are all degree-2 equations,

A x^{2} + B x y + C y^{2} + D x + E y + F = 0.

Proposition 8.14 (The discriminant identifies the conic). For a nondegenerate conic, the discriminant $B^{2} - 4 A C$ tells you which type it is: negative $\Rightarrow$ ellipse (or circle), zero $\Rightarrow$ parabola, positive $\Rightarrow$ hyperbola.

Standard forms (after translation to center/vertex $(h, k)$ and rotation to remove any $x y$ term):

Circle: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Parabola: $(x - h)^{2} = 4 p (y - k)$ , or the horizontal swap $(y - k)^{2} = 4 p (x - h)$
Ellipse: $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$
Hyperbola: $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$ , or the swapped form with the $y$ term positive

Each conic also has an equivalent distance definition in terms of foci or a focus and a directrix.

Definition 8.15 (Focal definitions of the conics).

Parabola: the set of points equidistant from a fixed focus and a fixed line, the directrix. Here $p$ is the distance from the vertex to the focus (and also to the directrix).
Ellipse: the set of points whose distances to two foci have a constant sum. The foci lie on the major axis at distance $c$ from the center, with $c^{2} = a^{2} - b^{2}$ (using the larger denominator as $a^{2}$ ).
Hyperbola: the set of points whose distances to two foci have a constant difference. The foci lie on the transverse axis with $c^{2} = a^{2} + b^{2}$ .

Intuition. All these curves are "conic" because you can get them by slicing an ice-cream cone (a double cone) with one flat cut at different tilts: straight across gives a circle, a gentle tilt an ellipse, a cut parallel to the side a parabola, and a steep cut through both cones a hyperbola. The algebra unifies all four as quadratic equations.

Example 8.16 (Identifying a conic and its features). Identify and give the key features of $9 x^{2} + 4 y^{2} - 36 = 0$ .

Solution. Move the constant and divide through by $36$ to get $1$ on the right: $\frac{9 x ^{2}}{36} + \frac{4 y ^{2}}{36} = 1$ , i.e. $\frac{x ^{2}}{4} + \frac{y ^{2}}{9} = 1$ . Both squared terms are positive with different denominators, so this is an ellipse centered at $(0, 0)$ . Compare denominators: $9 > 4$ , so the larger denominator sits under $y^{2}$ . Thus $a^{2} = 9$ (so $a = 3$ ) belongs to the $y$ -direction — the major axis is vertical — and $b^{2} = 4$ (so $b = 2$ ) is the horizontal semi-axis. Vertices (ends of the major axis) at $(0, \pm 3)$ ; co-vertices at $(\pm 2, 0)$ . Foci on the major (vertical) axis at $(0, \pm c)$ with $c = a^{2} - b^{2} = 9 - 4 = 5$ . Sanity check: plug the vertex $(0, 3)$ into the original equation: $9 (0) + 4 (9) - 36 = 36 - 36 = 0$ ✓.□

Example 8.17 (Completing the square to reach standard form). Rewrite $x^{2} + y^{2} - 6 x + 4 y - 3 = 0$ in standard form and identify it.

Solution. Group $x$ and $y$ terms: $(x^{2} - 6 x) + (y^{2} + 4 y) = 3$ . Complete the square in $x$ : half of $- 6$ is $- 3$ , and $(- 3)^{2} = 9$ , so add $9$ . Complete the square in $y$ : half of $4$ is $2$ , and $2^{2} = 4$ , so add $4$ . Add the same amounts to the right side to keep the equation balanced: $(x^{2} - 6 x + 9) + (y^{2} + 4 y + 4) = 3 + 9 + 4$ . Factor: $(x - 3)^{2} + (y + 2)^{2} = 16$ . This matches $(x - h)^{2} + (y - k)^{2} = r^{2}$ , a circle with center $(3, - 2)$ and radius $r = 16 = 4$ . Sanity check: the point $(7, - 2)$ lies $4$ units right of the center, so it should be on the circle. Plug into the original: $7^{2} + (- 2)^{2} - 6 (7) + 4 (- 2) - 3 = 49 + 4 - 42 - 8 - 3 = 0$ ✓.□

Common Mistakes

Conic identification from equation: The sign of B²-4AC tells parabola/ellipse/hyperbola (degenerate cases aside); students often look only at signs of A and C.
Conic standard form completion: For translated conics, students forget to complete the square or divide correctly after moving constants.
a>b or b>a in ellipse: The larger denominator is the semi-major axis; vertices on that axis.
For hyperbola the transverse axis: The one with the positive term (difference for the other).
Foci vs center: c² = |a² - b²| for ellipse/hyperbola; foci offset from center by c along major/transverse.
Parabola p: Distance from vertex to focus/directrix is p; 4p in the equation.

Exercises for §8.5 →

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter