Chapter 9: Sequences & Series
Sequences and series form the bridge between algebra and calculus. In this chapter, you will learn how to identify patterns in ordered lists of numbers, compute their sums using compact formulas, and apply powerful tools like the Binomial Theorem and mathematical induction. These ideas appear throughout higher mathematics, from financial modeling to physics and computer science.
Table of Contents
9.1 Introduction to Sequences
A sequence can be defined in two ways:
- Explicit formula: A formula that gives the $n$th term directly as a function of $n$. For example, $a_n = 3n + 1$ produces the sequence $4, 7, 10, 13, \ldots$
- Recursive formula: A formula that defines each term using one or more previous terms, together with an initial term. For example, $a_1 = 2,\; a_n = a_{n-1} + 5$ produces $2, 7, 12, 17, \ldots$
Find the first five terms of the sequence defined by $a_n = \dfrac{(-1)^{n+1}}{n^2}$.
Solution:
- $a_1 = \dfrac{(-1)^2}{1} = 1$
- $a_2 = \dfrac{(-1)^3}{4} = -\dfrac{1}{4}$
- $a_3 = \dfrac{(-1)^4}{9} = \dfrac{1}{9}$
- $a_4 = \dfrac{(-1)^5}{16} = -\dfrac{1}{16}$
- $a_5 = \dfrac{(-1)^6}{25} = \dfrac{1}{25}$
The first five terms are $1,\; -\dfrac{1}{4},\; \dfrac{1}{9},\; -\dfrac{1}{16},\; \dfrac{1}{25}$.
Write an explicit formula for the sequence $3, 7, 11, 15, 19, \ldots$
Solution: The terms increase by $4$ each time, so this is arithmetic with first term $a_1 = 3$ and common difference $d = 4$. The explicit formula is:
$$a_n = 3 + (n-1) \cdot 4 = 4n - 1$$9.2 Arithmetic Sequences and Series
where $a_1$ is the first term and $d$ is the common difference.
To find the common difference, subtract any term from the next: $d = a_{n+1} - a_n$. If this value is constant for all consecutive pairs, the sequence is arithmetic.
An arithmetic sequence has $a_1 = 5$ and $d = -3$. Find $a_{20}$.
Solution:
$$a_{20} = 5 + (20 - 1)(-3) = 5 + 19(-3) = 5 - 57 = -52$$Sum of an Arithmetic Series
The first form is intuitive: the average of the first and last term, multiplied by the number of terms. Legend has it that young Carl Friedrich Gauss discovered this idea when asked to add the integers from $1$ to $100$: he paired $1 + 100$, $2 + 99$, $3 + 98$, and so on, getting $50$ pairs each summing to $101$, for a total of $5050$.
Find the sum $5 + 8 + 11 + 14 + \cdots + 302$.
Solution: Here $a_1 = 5$, $d = 3$, and $a_n = 302$. First find $n$:
$$302 = 5 + (n-1)(3) \implies 297 = 3(n-1) \implies n = 100$$Now compute the sum:
$$S_{100} = \frac{100}{2}(5 + 302) = 50 \times 307 = 15{,}350$$9.3 Geometric Sequences and Series
where $a_1$ is the first term and $r$ is the common ratio.
To find the common ratio, divide any term by the previous term: $r = \dfrac{a_{n+1}}{a_n}$. If this ratio is constant for all consecutive pairs, the sequence is geometric.
A geometric sequence has $a_1 = 3$ and $r = 2$. Find $a_8$.
Solution:
$$a_8 = 3 \cdot 2^{8-1} = 3 \cdot 2^7 = 3 \cdot 128 = 384$$Sum of a Finite Geometric Series
To derive this, write $S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}$. Multiply both sides by $r$ and subtract to eliminate most terms, leaving $S_n(1 - r) = a_1(1 - r^n)$.
Find the sum $\displaystyle\sum_{k=1}^{8} 5 \cdot 3^{k-1}$.
Solution: Here $a_1 = 5$, $r = 3$, and $n = 8$.
$$S_8 = 5 \cdot \frac{1 - 3^8}{1 - 3} = 5 \cdot \frac{1 - 6561}{-2} = 5 \cdot \frac{-6560}{-2} = 5 \cdot 3280 = 16{,}400$$9.4 Infinite Geometric Series
What happens if we add up infinitely many terms of a geometric sequence? If the common ratio is small enough in absolute value, the partial sums approach a finite limit, and we say the infinite series converges.
If $|r| \geq 1$, the series diverges (the partial sums grow without bound or oscillate).
The condition $|r| < 1$ means the terms shrink toward zero, allowing the sum to settle on a finite value. When $|r| \geq 1$, each successive term is at least as large as the previous one, so the sum cannot converge.
Find the sum of $12 + 4 + \dfrac{4}{3} + \dfrac{4}{9} + \cdots$
Solution: The first term is $a_1 = 12$ and the common ratio is $r = \dfrac{4}{12} = \dfrac{1}{3}$. Since $|r| = \dfrac{1}{3} < 1$, the series converges:
$$S = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 12 \cdot \frac{3}{2} = 18$$Express the repeating decimal $0.272727\ldots$ as a fraction.
Solution: Write as a geometric series:
$$0.272727\ldots = 0.27 + 0.0027 + 0.000027 + \cdots = \frac{27}{100} + \frac{27}{10000} + \frac{27}{1000000} + \cdots$$This is geometric with $a_1 = \dfrac{27}{100}$ and $r = \dfrac{1}{100}$. Since $|r| < 1$:
$$S = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}$$9.5 Sigma Notation and Properties of Summation
reads "the sum of $a_k$ from $k = m$ to $k = n$." Here $k$ is the index of summation, $m$ is the lower limit, and $n$ is the upper limit.
Properties of Summation
- $\displaystyle\sum_{k=1}^{n} c \cdot a_k = c \cdot \sum_{k=1}^{n} a_k$ (constant multiple)
- $\displaystyle\sum_{k=1}^{n} (a_k \pm b_k) = \sum_{k=1}^{n} a_k \pm \sum_{k=1}^{n} b_k$ (sum/difference)
- $\displaystyle\sum_{k=1}^{n} c = nc$ (sum of a constant)
Useful Closed-Form Formulas
- $\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$
- $\displaystyle\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
- $\displaystyle\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$
Evaluate $\displaystyle\sum_{k=1}^{50} (3k + 2)$.
Solution: Use the summation properties:
$$\sum_{k=1}^{50} (3k + 2) = 3\sum_{k=1}^{50} k + \sum_{k=1}^{50} 2 = 3 \cdot \frac{50 \cdot 51}{2} + 2 \cdot 50 = 3 \cdot 1275 + 100 = 3825 + 100 = 3925$$9.6 Mathematical Induction
Mathematical induction is a proof technique used to establish that a statement $P(n)$ holds for all positive integers $n$. It works like a chain of dominoes: if the first domino falls, and each domino knocks down the next, then all dominoes fall.
- Base Case: Show that $P(1)$ is true.
- Inductive Step: Assume $P(k)$ is true for some arbitrary positive integer $k$ (the inductive hypothesis). Then show that $P(k+1)$ must also be true.
If both steps are completed, then $P(n)$ is true for all integers $n \geq 1$.
Prove that $\displaystyle\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ for all positive integers $n$.
Proof:
Base Case ($n = 1$): LHS $= 1$. RHS $= \dfrac{1 \cdot 2}{2} = 1$. True.
Inductive Step: Assume the formula holds for $n = k$, meaning $\displaystyle\sum_{j=1}^{k} j = \frac{k(k+1)}{2}$.
We must show it holds for $n = k + 1$:
$$\sum_{j=1}^{k+1} j = \sum_{j=1}^{k} j + (k+1) = \frac{k(k+1)}{2} + (k+1)$$ $$= (k+1)\left(\frac{k}{2} + 1\right) = (k+1) \cdot \frac{k+2}{2} = \frac{(k+1)(k+2)}{2}$$This is exactly the formula with $n = k+1$. By induction, the formula holds for all $n \geq 1$. $\square$
Prove that $6^n - 1$ is divisible by $5$ for all positive integers $n$.
Proof:
Base Case ($n = 1$): $6^1 - 1 = 5$, which is divisible by $5$. True.
Inductive Step: Assume $6^k - 1 \equiv 0 \pmod{5}$ for some $k$, so $6^k - 1 = 5m$ for some integer $m$. Then:
$$6^{k+1} - 1 = 6 \cdot 6^k - 1 = 6(5m + 1) - 1 = 30m + 6 - 1 = 30m + 5 = 5(6m + 1)$$Since $5(6m+1)$ is divisible by $5$, the statement holds for $k+1$. By induction, $6^n - 1$ is divisible by $5$ for all $n \geq 1$. $\square$
9.7 The Binomial Theorem and Pascal's Triangle
Binomial Coefficients
This counts the number of ways to choose $k$ items from a set of $n$ items without regard to order.
The connection to counting is fundamental: $\binom{n}{k}$ appears in the Binomial Theorem precisely because expanding $(a + b)^n$ involves choosing which of the $n$ factors contribute $a$ and which contribute $b$. To get the term $a^{n-k}b^k$, you must choose $k$ of the $n$ factors to supply the $b$, which can be done in $\binom{n}{k}$ ways.
Pascal's Triangle
Pascal's Triangle arranges the binomial coefficients in a triangular array. Each entry is the sum of the two entries directly above it:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Row $n$ contains the coefficients $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$, starting from row $0$ at the top.
This explains why each entry in Pascal's Triangle is the sum of the two entries above it.
The Binomial Theorem
Key observations about the expansion of $(a+b)^n$:
- There are $n + 1$ terms in the expansion.
- The exponents of $a$ decrease from $n$ to $0$, while the exponents of $b$ increase from $0$ to $n$.
- In each term, the sum of the exponents of $a$ and $b$ equals $n$.
- The coefficients are symmetric: $\binom{n}{k} = \binom{n}{n-k}$.
Expand $(2x + 3)^4$.
Solution: Apply the Binomial Theorem with $a = 2x$, $b = 3$, $n = 4$:
$$(2x+3)^4 = \sum_{k=0}^{4} \binom{4}{k}(2x)^{4-k}(3)^k$$- $k=0$: $\binom{4}{0}(2x)^4(3)^0 = 1 \cdot 16x^4 \cdot 1 = 16x^4$
- $k=1$: $\binom{4}{1}(2x)^3(3)^1 = 4 \cdot 8x^3 \cdot 3 = 96x^3$
- $k=2$: $\binom{4}{2}(2x)^2(3)^2 = 6 \cdot 4x^2 \cdot 9 = 216x^2$
- $k=3$: $\binom{4}{3}(2x)^1(3)^3 = 4 \cdot 2x \cdot 27 = 216x$
- $k=4$: $\binom{4}{4}(2x)^0(3)^4 = 1 \cdot 1 \cdot 81 = 81$
Find the coefficient of $x^5$ in the expansion of $(x + 2)^8$.
Solution: The general term is $\binom{8}{k} x^{8-k} \cdot 2^k$. We need $8 - k = 5$, so $k = 3$.
$$\binom{8}{3} x^5 \cdot 2^3 = 56 \cdot x^5 \cdot 8 = 448x^5$$The coefficient of $x^5$ is $\mathbf{448}$.
9.8 Practice Problems
Test your understanding with these problems covering all topics from this chapter. Click "Show Solution" to check your work.
The 5th term of an arithmetic sequence is $19$ and the 12th term is $47$. Find the first term and the common difference.
Show Solution
We have $a_5 = a_1 + 4d = 19$ and $a_{12} = a_1 + 11d = 47$.
Subtracting the first equation from the second: $7d = 28$, so $d = 4$.
Substituting back: $a_1 + 4(4) = 19$, so $a_1 = 19 - 16 = 3$.
The first term is $a_1 = 3$ and the common difference is $d = 4$.
Find $\displaystyle\sum_{k=1}^{40} (2k + 5)$.
Show Solution
This is an arithmetic series with $a_1 = 2(1)+5 = 7$ and $a_{40} = 2(40)+5 = 85$.
$$S_{40} = \frac{40}{2}(7 + 85) = 20 \times 92 = 1840$$Alternatively: $\displaystyle\sum_{k=1}^{40}(2k+5) = 2\cdot\frac{40 \cdot 41}{2} + 5 \cdot 40 = 1640 + 200 = 1840$.
A geometric sequence has $a_3 = 20$ and $a_6 = 160$. Find the common ratio $r$ and the first term $a_1$.
Show Solution
We have $a_3 = a_1 r^2 = 20$ and $a_6 = a_1 r^5 = 160$.
Dividing: $\dfrac{a_6}{a_3} = \dfrac{a_1 r^5}{a_1 r^2} = r^3 = \dfrac{160}{20} = 8$, so $r = 2$.
From $a_1 r^2 = 20$: $a_1 \cdot 4 = 20$, so $a_1 = 5$.
Find the sum $\displaystyle\sum_{k=0}^{9} 6 \cdot \left(\frac{1}{2}\right)^k$.
Show Solution
Here $a_1 = 6$, $r = \dfrac{1}{2}$, and $n = 10$ (from $k=0$ to $k=9$).
$$S_{10} = 6 \cdot \frac{1 - (1/2)^{10}}{1 - 1/2} = 6 \cdot \frac{1 - 1/1024}{1/2} = 12 \cdot \frac{1023}{1024} = \frac{12276}{1024} = \frac{3069}{256} \approx 11.988$$Determine whether each series converges or diverges. If it converges, find the sum.
(a) $\displaystyle\sum_{n=1}^{\infty} 8 \cdot \left(-\frac{2}{3}\right)^{n-1}$ (b) $\displaystyle\sum_{n=0}^{\infty} \frac{5^n}{4^n}$
Show Solution
(a) Here $a_1 = 8$ and $r = -\dfrac{2}{3}$. Since $|r| = \dfrac{2}{3} < 1$, the series converges:
$$S = \frac{8}{1 - (-2/3)} = \frac{8}{5/3} = \frac{24}{5}$$(b) This is $\displaystyle\sum_{n=0}^{\infty} \left(\frac{5}{4}\right)^n$, which is geometric with $r = \dfrac{5}{4}$. Since $|r| = \dfrac{5}{4} > 1$, the series diverges.
Express $0.5\overline{18} = 0.518181818\ldots$ as a fraction.
Show Solution
Write $0.518181818\ldots = 0.5 + 0.018181818\ldots$
The repeating part is $0.0\overline{18} = \dfrac{18}{1000} + \dfrac{18}{100000} + \cdots$, a geometric series with $a_1 = \dfrac{18}{1000}$ and $r = \dfrac{1}{100}$:
$$0.0\overline{18} = \frac{18/1000}{1 - 1/100} = \frac{18/1000}{99/100} = \frac{18}{990} = \frac{1}{55}$$Therefore: $0.5\overline{18} = \dfrac{1}{2} + \dfrac{1}{55} = \dfrac{55 + 2}{110} = \dfrac{57}{110}$.
Evaluate $\displaystyle\sum_{k=1}^{20} k^2$.
Show Solution
Using the standard formula:
$$\sum_{k=1}^{20} k^2 = \frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870$$Prove by induction that $\displaystyle\sum_{k=1}^{n} (2k - 1) = n^2$ for all positive integers $n$.
Show Solution
Base Case ($n = 1$): LHS $= 2(1) - 1 = 1$. RHS $= 1^2 = 1$. True.
Inductive Step: Assume $\displaystyle\sum_{j=1}^{k}(2j-1) = k^2$. Show the formula holds for $k+1$:
$$\sum_{j=1}^{k+1}(2j-1) = \sum_{j=1}^{k}(2j-1) + [2(k+1)-1] = k^2 + (2k+1) = (k+1)^2$$This is the formula with $n = k+1$. By mathematical induction, $\displaystyle\sum_{k=1}^{n}(2k-1) = n^2$ for all $n \geq 1$. $\square$
Expand $(x - 2)^5$ using the Binomial Theorem.
Show Solution
Apply the Binomial Theorem with $a = x$, $b = -2$, $n = 5$:
$$(x-2)^5 = \sum_{k=0}^{5}\binom{5}{k}x^{5-k}(-2)^k$$- $k=0$: $\binom{5}{0}x^5(-2)^0 = x^5$
- $k=1$: $\binom{5}{1}x^4(-2)^1 = -10x^4$
- $k=2$: $\binom{5}{2}x^3(-2)^2 = 40x^3$
- $k=3$: $\binom{5}{3}x^2(-2)^3 = -80x^2$
- $k=4$: $\binom{5}{4}x^1(-2)^4 = 80x$
- $k=5$: $\binom{5}{5}x^0(-2)^5 = -32$
Find the term containing $x^3$ in the expansion of $(3x + 4)^6$.
Show Solution
The general term is $\binom{6}{k}(3x)^{6-k}(4)^k$. For the $x^3$ term, we need $6 - k = 3$, so $k = 3$.
$$\binom{6}{3}(3x)^3(4)^3 = 20 \cdot 27x^3 \cdot 64 = 20 \cdot 27 \cdot 64 \cdot x^3 = 34{,}560x^3$$A pizza shop offers 10 toppings. How many different pizzas can be made with exactly 4 toppings?
Show Solution
The order of toppings does not matter, so we use combinations:
$$\binom{10}{4} = \frac{10!}{4! \cdot 6!} = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = \frac{5040}{24} = 210$$There are $\mathbf{210}$ different 4-topping pizzas.
A ball is dropped from a height of $10$ meters. Each time it bounces, it reaches $\dfrac{3}{5}$ of the previous height. Find the total vertical distance traveled by the ball.
Show Solution
The ball falls $10$ m initially. It then bounces up $10 \cdot \dfrac{3}{5} = 6$ m and falls $6$ m, then bounces up $6 \cdot \dfrac{3}{5} = 3.6$ m and falls $3.6$ m, and so on.
Total distance $= 10 + 2(6 + 3.6 + 2.16 + \cdots) = 10 + 2 \displaystyle\sum_{n=1}^{\infty} 10 \cdot \left(\frac{3}{5}\right)^n$.
The sum $\displaystyle\sum_{n=1}^{\infty} 10 \cdot \left(\frac{3}{5}\right)^n = 10 \cdot \frac{3/5}{1 - 3/5} = 10 \cdot \frac{3/5}{2/5} = 10 \cdot \frac{3}{2} = 15$.
Total distance $= 10 + 2(15) = 10 + 30 = \mathbf{40}$ meters.