Honors Pre-Calculus Unit 11 Review: Sequences, Probability & Counting Theory

A sequence is an ordered list of numbers, called terms, that follows a rule. The notation $a_n$ means the term in position $n$, so $a_1$ is the first term and $a_{10}$ is the tenth. A finite sequence stops after some number of terms, and an infinite sequence keeps going.

There are two ways to describe the rule. An explicit formula gives $a_n$ directly from $n$. With $a_n = 3n - 1$, plug in $n = 10$ and you get $a_{10} = 29$ right away.

A recursive formula defines a term using earlier terms. For example, $a_n = a_{n-1} + 3$ with $a_1 = 2$ gives the same sequence 2, 5, 8, 11, .... To find the tenth term by applying this rule repeatedly, calculate the earlier terms first. A recursive formula always needs a starting value, or it can't produce anything. The Fibonacci sequence 1, 1, 2, 3, 5, 8, ... is recursive with two starting values, since each term is the sum of the previous two.

Factorials show up here and again later in the unit. The notation $n!$ means $n \cdot (n-1) \cdots 2 \cdot 1$, so $4! = 24$. By definition, $0! = 1$.

A sequence is a function whose inputs are positive integers, so its graph is a set of separate dots, not a connected curve.

An arithmetic sequence adds the same number, the common difference $d$, to get from each term to the next. You find it by subtracting consecutive terms, $d = a_{n+1} - a_n$. If the differences aren't all equal, the sequence isn't arithmetic. A positive $d$ makes the sequence increase, and a negative $d$ makes it decrease.

The recursive form is $a_n = a_{n-1} + d$. The explicit form is $a_n = a_1 + (n-1)d$. The $n - 1$ is there because the first term has had zero differences added to it.

Arithmetic sequences are linear functions in disguise. For $a_n = 5 + (n-1)(4)$, simplifying gives $a_n = 4n + 1$, a line with slope 4. The common difference plays the role of the slope, which is why the graph of the terms is a set of dots along a straight line.

Worked example. An arithmetic sequence has $a_4 = 17$ and $a_{10} = 41$. Find $a_{30}$. Going from term 4 to term 10 takes 6 steps, so $6d = 41 - 17 = 24$ and $d = 4$. Work backward 3 steps from $a_4$ to get $a_1 = 17 - 3(4) = 5$. Then $a_{30} = 5 + (30 - 1)(4) = 5 + 116 = 121$.

A geometric sequence multiplies each term by the same number, the common ratio $r$, to get the next term. You can find it by dividing consecutive terms, $r = \dfrac{a_{n+1}}{a_n}$, when $a_n \neq 0$. The recursive form is $a_n = r \cdot a_{n-1}$, and the explicit form is $a_n = a_1 r^{n-1}$. If $r = 0$, use $a_1$ for the first term and 0 for every later term.

The exponent is $n - 1$, not $n$. The first term has been multiplied by $r$ zero times, and $r^0 = 1$. For example, with $a_1 = 2$ and $r = 3$, $a_{10} = 2 \cdot 3^9 = 2 \cdot 19{,}683 = 39{,}366$.

The value of $r$ controls the behavior of a geometric sequence with a nonzero first term.

• If $|r| > 1$, the terms grow in magnitude, like 2, 6, 18, 54.
• If $|r| < 1$, the terms approach 0, like 8, 4, 2, 1. The sequence converges to 0.
• If $r$ is negative, the signs alternate. With $r = -2$, you get 1, -2, 4, -8.

Geometric sequences with positive ratios follow exponential patterns and can model compound interest, population growth and half-life decay. Arithmetic sequences add a fixed amount, while geometric sequences multiply by a fixed factor. A negative ratio produces alternating signs rather than an ordinary real exponential function.

A series is the sum of the terms of a sequence. Sigma notation writes it compactly. In $\sum_{k=1}^{5} 2k$, the index $k$ starts at the lower limit 1, stops at the upper limit 5, and you add the outputs to get $2 + 4 + 6 + 8 + 10 = 30$. The sum of the first $n$ terms is called a partial sum, $S_n$.

Three sum formulas do most of the work in this topic.

• Arithmetic series use $S_n = \frac{n}{2}(a_1 + a_n)$, which is the average of the first and last terms times the number of terms.
• Finite geometric series use $S_n = \dfrac{a_1(1 - r^n)}{1 - r}$ for $r \neq 1$.
• Infinite geometric series use $S_\infty = \dfrac{a_1}{1 - r}$ when $|r| < 1$. If $a_1 \neq 0$ and $|r| \geq 1$, the terms do not approach 0, so the series diverges. If $a_1 = 0$, every term is 0 and the sum is 0.

Worked example. Add the integers from 1 to 100. This is arithmetic with $n = 100$, $a_1 = 1$ and $a_{100} = 100$, so $S_{100} = \frac{100}{2}(1 + 100) = 50(101) = 5050$.

Worked example. Find $12 + 4 + \frac{4}{3} + \cdots$. Here $a_1 = 12$ and $r = \frac{1}{3}$. Since $|r| < 1$, the sum exists, and $S_\infty = \dfrac{12}{1 - \frac{1}{3}} = \dfrac{12}{2/3} = 18$.

Counting principles let you find how many outcomes exist without listing them. The addition principle handles "or" choices that can't happen together. If you pick one item from 5 shirts or 3 hats, you have $5 + 3 = 8$ choices. The multiplication principle handles "and" choices made in stages. A shirt and a hat give $5 \times 3 = 15$ outfits.

Permutations count arrangements where order matters. Arranging all $n$ distinct objects gives $n!$ ways, and arranging $r$ of them gives $P(n, r) = \dfrac{n!}{(n-r)!}$.

Combinations count selections where order doesn't matter, using $C(n, r) = \binom{n}{r} = \dfrac{n!}{r!(n-r)!}$. Each group of $r$ items can be ordered $r!$ ways, so $C(n, r) = \dfrac{P(n, r)}{r!}$.

Worked example. A club has 10 members. Electing a president, vice president and secretary means the positions are different, so order matters, and $P(10, 3) = 10 \cdot 9 \cdot 8 = 720$. Picking a 3-person committee means order doesn't matter, so $C(10, 3) = \dfrac{720}{3!} = 120$.

Two extra cases come up often. When letters repeat, divide by the factorial of each repeat count, so BOOK has $\dfrac{4!}{2!} = 12$ arrangements. A set with $n$ elements has $2^n$ subsets, counting the empty set, because each element is either in or out.

For a nonnegative integer $n$, the binomial theorem expands $(x + y)^n$ without multiplying it out by hand:

$(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$

The expansion has $n + 1$ terms. The power on $x$ counts down from $n$, the power on $y$ counts up from 0, and in every term the two powers add to $n$. For example, $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

The coefficients $\binom{n}{k}$ are the same combinations from counting. You can also read them off Pascal's triangle. After the top 1, each entry is the sum of the two entries above it, with missing entries treated as 0. Row 4 (counting the top row as row 0) is 1, 4, 6, 4, 1, which gives the coefficients of $(x + y)^4$.

With a subtraction like $(a - b)^n$, treat it as $(a + (-b))^n$ and the signs will alternate.

Worked example. Find the 4th term of $(3x - 2y)^6$. The 4th term uses $k = 3$, because counting starts at $k = 0$. The coefficient is $\binom{6}{3} = 20$. The term is $20(3x)^3(-2y)^3 = 20 \cdot 27x^3 \cdot (-8y^3) = -4320x^3y^3$.

Probability measures how likely an event is on a scale from 0 to 1. An impossible event has probability 0, and a certain event has probability 1. The sample space is the set of all possible outcomes, and an event is a subset of it. When every outcome is equally likely, $P(E) = \dfrac{\text{favorable outcomes}}{\text{total outcomes}}$. On a fair die, $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$. Counting methods from 11.5 often supply both numbers. The chance of drawing 2 aces from a deck is $\dfrac{\binom{4}{2}}{\binom{52}{2}} = \dfrac{6}{1326} = \dfrac{1}{221}$.

A few rules handle most problems.

• The complement rule says $P(\text{not } A) = 1 - P(A)$.
• For "A or B," use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. You subtract the overlap so it isn't counted twice. If the events are mutually exclusive, the overlap is 0.
• For independent events, where one doesn't affect the other, $P(A \cap B) = P(A) \cdot P(B)$.

Worked example. Find the probability of at least one 6 in four rolls of a fair die. Counting "at least one" directly is messy, so use the complement. The chance of no 6 on one roll is $\frac{5}{6}$, and the rolls are independent, so $P(\text{no 6s}) = \left(\frac{5}{6}\right)^4 = \frac{625}{1296} \approx 0.482$. That makes $P(\text{at least one 6}) \approx 1 - 0.482 = 0.518$.