Precalculus · Chapter 12

Sequences and Series

SECTION 12.1

Sequences and Summation Notation

— Sequence

A sequence is a function $ a $ whose domain is the set of natural numbers. The terms of the sequence are the function values $ a(1), a(2), a(3), \dots, a(n), \dots $. We usually write $ a_n $ instead of $ a(n) $, so the terms are written $ a_1, a_2, a_3, \dots, a_n, \dots $. The number $ a_1 $ is the first term, $ a_2 $ the second term, and $ a_n $ the nth term. A sequence can be thought of as a list of numbers written in a specific order.

— Recursively Defined Sequence

A sequence is recursive when the nth term depends on some or all of the terms preceding it. Instead of a direct formula for $ a_n $, a recursion relation (e.g. $ a_n = a_{n-1} + 5 $) plus a starting value defines every term.

— Partial Sums

Adding the terms of a sequence produces its sequence of partial sums. $ S_1 $ is the first partial sum, $ S_2 $ the second, and $ S_n $ the nth partial sum. The sequence $ S_1, S_2, S_3, \dots $ is called the sequence of partial sums.

— Sigma (Summation) Notation

Sigma notation uses the Greek capital letter $ \Sigma $ to abbreviate a sum. The letter $ k $ is the index of summation (or summation variable): we replace $ k $ by the integers from the lower limit to the upper limit and add the results.

— nth Term (Direct Formula)

\[ a_n = f(n) \] A direct formula gives any term from its index. *Example:* $ a_n = 2n $ generates $ 2, 4, 6, 8, \dots $

— Partial Sums of a Sequence

\[ S_n = a_1 + a_2 + a_3 + \cdots + a_n = \sum_{k=1}^{n} a_k \]

— Sigma Notation

\[ \sum_{k=1}^{n} a_k = a_1 + a_2 + a_3 + \cdots + a_n \]

— Properties of Sums

\[ \sum_{k=1}^{n}(a_k + b_k) = \sum_{k=1}^{n}a_k + \sum_{k=1}^{n}b_k \] \[ \sum_{k=1}^{n}(a_k - b_k) = \sum_{k=1}^{n}a_k - \sum_{k=1}^{n}b_k \] \[ \sum_{k=1}^{n} c\,a_k = c\left(\sum_{k=1}^{n}a_k\right) \]

EXAMPLE 1 Finding the Terms of a Sequence

Find the first five terms and the 100th term of the sequence $ a_n = 2n - 1 $.

Solution

Substitute $ n = 1, 2, 3, 4, 5 $: $ a_1 = 1,\ a_2 = 3,\ a_3 = 5,\ a_4 = 7,\ a_5 = 9 $.

The first five terms are $ 1, 3, 5, 7, 9 $.

Substitute $ n = 100 $: $ a_{100} = 2(100) - 1 = 199 $.

EXAMPLE 2 Finding the nth Term of a Sequence

Find the nth term of the sequence $ \tfrac{1}{2}, \tfrac{3}{4}, \tfrac{5}{6}, \tfrac{7}{8}, \dots $

Solution

The numerators are the odd numbers, which have the form $ 2n - 1 $.

The denominators are the even numbers, which have the form $ 2n $.

Therefore \[ a_n = \frac{2n-1}{2n} \]

EXAMPLE 3 Finding the Terms of a Recursively Defined Sequence

A sequence is defined recursively by $ a_1 = 1 $ and $ a_n = 3(a_{n-1} + 2) $. Find the first five terms.

Solution

$ a_2 = 3(a_1 + 2) = 3(1 + 2) = 9 $

$ a_3 = 3(a_2 + 2) = 3(9 + 2) = 33 $

$ a_4 = 3(a_3 + 2) = 3(33 + 2) = 105 $

$ a_5 = 3(a_4 + 2) = 3(105 + 2) = 321 $

The first five terms are $ 1, 9, 33, 105, 321 $.

EXAMPLE 4 The Fibonacci Sequence

Find the first 11 terms of the sequence defined by $ F_1 = 1,\ F_2 = 1, $ and $ F_n = F_{n-1} + F_{n-2} $.

Solution

Each term is the sum of the two preceding terms: $ F_3 = 1+1 = 2 $, $ F_4 = 2+1 = 3 $, $ F_5 = 3+2 = 5 $, and so on.

First 11 terms: $ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 $.

EXAMPLE 5 Finding the Partial Sums of a Sequence

Find the first four partial sums and the nth partial sum of $ a_n = \tfrac{1}{2^n} $.

Solution

$ S_1 = \tfrac{1}{2} $

$ S_2 = \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4} $

$ S_3 = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} = \tfrac{7}{8} $

$ S_4 = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{16} = \tfrac{15}{16} $

In each sum the numerator is one less than the denominator (a power of 2): \[ S_n = \frac{2^n - 1}{2^n} = 1 - \frac{1}{2^n} \]

EXAMPLE 7 Sigma Notation

Find each sum: (a) $ \sum_{k=1}^{5} k^2 $ (b) $ \sum_{j=3}^{5} \tfrac{1}{j} $

Solution

(a) $ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 $

(b) $ \tfrac{1}{3} + \tfrac{1}{4} + \tfrac{1}{5} = \tfrac{47}{60} $

EXAMPLE 8 Writing Sums in Sigma Notation

Write $ 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 $ using sigma notation.

Solution

Each term is a cube of the index, from 1 to 7: \[ \sum_{k=1}^{7} k^3 \]

Instructor Note

Instructor note — to be added.

SECTION 12.2

Arithmetic Sequences

— Arithmetic Sequence

An arithmetic sequence is generated by starting with a number $ a $ and repeatedly adding a fixed constant $ d $. It has the form $ a,\ a+d,\ a+2d,\ a+3d,\ \dots $ The number $ a $ is the first term and $ d $ is the common difference — any two consecutive terms differ by $ d $. If $ d $ is negative, the terms decrease.

— Determining an Arithmetic Sequence

An arithmetic sequence is determined completely by its first term $ a $ and common difference $ d $. Knowing any two terms is enough to find a formula for the nth term. The points on the graph of an arithmetic sequence lie on a straight line of slope $ d $.

— nth Term of an Arithmetic Sequence

\[ a_n = a + (n-1)d \]

— Partial Sum of an Arithmetic Sequence (Form 1)

\[ S_n = \frac{n}{2}\left[\,2a + (n-1)d\,\right] \]

— Partial Sum of an Arithmetic Sequence (Form 2)

\[ S_n = n\left(\frac{a + a_n}{2}\right) \] The sum equals the average of the first and last terms, multiplied by the number of terms.

EXAMPLE 2 Finding Terms of an Arithmetic Sequence

Find the common difference, the nth term, and the 300th term of $ 13, 7, 1, -5, \dots $

Solution

First term: $ a = 13 $.

Common difference: $ d = 7 - 13 = -6 $.

nth term: $ a_n = 13 - 6(n-1) $.

$ a_{300} = 13 - 6(300 - 1) = -1781 $.

EXAMPLE 3 Finding Terms from Two Given Terms

The 11th term of an arithmetic sequence is 52 and the 19th term is 92. Find the 1000th term.

Solution

Set up equations using $ a_n = a + (n-1)d $: $ a_{11} = a + 10d = 52 $ and $ a_{19} = a + 18d = 92 $.

Solve the system: subtracting gives $ 8d = 40 $, so $ d = 5 $ and $ a = 2 $.

nth term: $ a_n = 2 + 5(n-1) $.

$ a_{1000} = 2 + 5(1000 - 1) = 4997 $.

EXAMPLE 4 Finding a Partial Sum

Find the sum of the first 50 odd numbers.

Solution

The odd numbers form an arithmetic sequence with $ a = 1 $ and $ d = 2 $.

The 50th odd number: $ a_{50} = 2(50) - 1 = 99 $.

Apply Form 2: $ S_{50} = 50\left(\frac{1 + 99}{2}\right) = 50 \cdot 50 = 2500 $.

EXAMPLE 5 Finding a Partial Sum

Find $ 3 + 7 + 11 + 15 + \cdots + 159 $.

Solution

Here $ a = 3 $ and $ d = 4 $, so $ a_n = 3 + 4(n-1) $.

Find which term equals 159: $ 159 = 3 + 4(n-1) \Rightarrow n = 40 $.

Apply Form 1: $ S_{40} = \frac{40}{2}\left[2(3) + 4(40 - 1)\right] = 3240 $.

EXAMPLE 6 Seating Capacity of an Amphitheater

An amphitheater has 50 rows: 30 seats in the first row, 32 in the second, 34 in the third, and so on. Find the total number of seats.

Solution

Seats per row form an arithmetic sequence with $ a = 30 $, $ d = 2 $, $ n = 50 $.

$ S_{50} = \frac{50}{2}\left[2(30) + 49(2)\right] = 3950 $ seats.

EXAMPLE 7 Finding the Number of Terms

How many terms of $ 5, 7, 9, \dots $ must be added to get 572?

Solution

Here $ a = 5 $, $ d = 2 $, $ S_n = 572 $. Substitute into Form 1: $ 572 = \frac{n}{2}\left[2(5) + (n-1)2\right] $.

Simplify: $ 572 = 5n + n(n-1) \Rightarrow 0 = n^2 + 4n - 572 $.

Factor: $ 0 = (n - 22)(n + 26) $, giving $ n = 22 $ (rejecting negative).

So 22 terms must be added.

Instructor Note

Instructor note — to be added.

SECTION 12.3

Geometric Sequences

— Geometric Sequence

A geometric sequence is generated by starting with a number $ a $ and repeatedly multiplying by a fixed nonzero constant $ r $. It has the form $ a, ar, ar^2, ar^3, ar^4, \dots $ The number $ a $ is the first term and $ r $ is the common ratio — the ratio of any two consecutive terms. If $ 0 < |r| < 1 $ the terms decrease in magnitude; if $ |r| > 1 $ they increase. The points lie on the graph of an exponential function.

— Infinite Series

An infinite series is an expression of the form $ \sum_{k=1}^{\infty} a_k = a_1 + a_2 + a_3 + \cdots $, where the addition continues indefinitely. If the partial sums $ S_n $ approach a finite number $ S $ as $ n $ grows large, the series converges and $ S $ is its sum. Otherwise the series diverges.

— Infinite Geometric Series

An infinite geometric series has the form $ a + ar + ar^2 + ar^3 + \cdots $. When $ |r| < 1 $, $ r^n $ approaches 0 as $ n $ grows large, so the partial sums approach a finite limit and the series converges. When $ |r| \geq 1 $, it diverges.

— nth Term of a Geometric Sequence

\[ a_n = a\,r^{\,n-1} \]

— Partial Sum of a Geometric Sequence

\[ S_n = a\,\frac{1 - r^{\,n}}{1 - r}, \qquad r \neq 1 \]

— Sum of an Infinite Geometric Series

\[ S = \frac{a}{1 - r}, \qquad |r| < 1 \] If $ |r| \geq 1 $, the series diverges.

EXAMPLE 2 Finding Terms of a Geometric Sequence

Find the common ratio, the nth term, and the eighth term of $ 5, 15, 45, 135, \dots $

Solution

First term: $ a = 5 $.

Common ratio: $ r = \frac{45}{15} = 3 $.

nth term: $ a_n = 5(3)^{n-1} $.

$ a_8 = 5(3)^{8-1} = 5 \cdot 2187 = 10{,}935 $.

EXAMPLE 3 Finding Terms from Two Given Terms

The third term of a geometric sequence is $ \tfrac{63}{4} $ and the sixth term is $ \tfrac{1701}{32} $. Find the fifth term.

Solution

Write $ a_3 = ar^2 = \tfrac{63}{4} $ and $ a_6 = ar^5 = \tfrac{1701}{32} $.

Divide: $ \frac{ar^5}{ar^2} = r^3 = \frac{1701/32}{63/4} = \frac{27}{8} $, so $ r = \tfrac{3}{2} $.

Substitute back: $ \tfrac{63}{4} = a\left(\tfrac{3}{2}\right)^2 \Rightarrow a = 7 $.

nth term: $ a_n = 7\left(\tfrac{3}{2}\right)^{n-1} $. Fifth term: $ a_5 = 7\left(\tfrac{3}{2}\right)^4 = \tfrac{567}{16} $.

EXAMPLE 4 Finding a Partial Sum

Find $ 1 + 4 + 16 + \cdots + 4096 $.

Solution

Here $ a = 1 $, $ r = 4 $, and $ a_n = 4^{n-1} $.

Since $ 4^6 = 4096 $, use $ n = 7 $: $ S_7 = 1 \cdot \frac{1 - 4^7}{1 - 4} = 5461 $.

EXAMPLE 5 Finding a Partial Sum (Sigma Form)

Find $ \sum_{k=1}^{6} 7\left(\tfrac{2}{3}\right)^{k-1} $.

Solution

This is the sixth partial sum of a geometric sequence with $ a = 7 $ and $ r = \tfrac{2}{3} $.

$ S_6 = 7 \cdot \frac{1 - (2/3)^6}{1 - 2/3} = \frac{931}{243} \approx 3.83 $.

EXAMPLE 6 Infinite Series

Determine convergence and, if convergent, the sum of: (a) $ 2 + \tfrac{2}{5} + \tfrac{2}{25} + \tfrac{2}{125} + \cdots $ (b) $ 1 + \tfrac{7}{5} + \left(\tfrac{7}{5}\right)^2 + \cdots $

Solution

(a) $ a = 2 $, $ r = \tfrac{1}{5} $. Since $ |r| < 1 $, it converges: $ S = \frac{2}{1 - 1/5} = \tfrac{5}{2} $.

(b) $ a = 1 $, $ r = \tfrac{7}{5} $. Since $ |r| > 1 $, the series diverges.

EXAMPLE 7 Writing a Repeated Decimal as a Fraction

Express $ 2.3\overline{51} $ as a fraction.

Solution

Write as $ \tfrac{23}{10} + \tfrac{51}{1000} + \tfrac{51}{100{,}000} + \cdots $.

After the first term, the rest is an infinite geometric series with $ a = \tfrac{51}{1000} $, $ r = \tfrac{1}{100} $.

Sum of the series: $ \frac{51/1000}{1 - 1/100} = \frac{51}{990} $.

Total: $ \tfrac{23}{10} + \tfrac{51}{990} = \tfrac{2328}{990} = \tfrac{388}{165} $.

Instructor Note

Instructor note — to be added.

SECTION 12.4

Mathematics of Finance

— Annuity

An annuity is a sum of money paid in regular equal payments (annually, semiannually, quarterly, monthly, etc.). The regular payment is called the periodic rent, denoted $ R $. The time period in which interest is compounded is assumed equal to the time between payments.

— Amount of an Annuity

The amount $ A_f $ of an annuity is the sum of all individual payments from the first payment until the last, together with all accumulated interest. It is the future (final) value of the payment stream.

— Present Value

The present value $ A_p $ (or discounted value) is the amount that, together with interest, would equal a future amount. Money received in the future is worth less than money now because of interest that could be earned in the meantime.

— Installment Buying

When buying by installment, the payments form an annuity whose present value is the amount of the loan. Solving for the payment size gives the installment-buying formula.

— Amount of an Annuity

\[ A_f = R\,\frac{(1 + i)^{n} - 1}{i} \] $ R $ = payment size, $ i $ = interest rate per period, $ n $ = number of payments.

— Present Value of an Annuity

\[ A_p = R\,\frac{1 - (1 + i)^{-n}}{i} \]

— Present Value of a Future Amount

\[ A_p = A_f\,(1 + i)^{-n} \]

— Installment Buying (Payment Size)

\[ R = \frac{i\,A_p}{1 - (1 + i)^{-n}} \]

EXAMPLE 1 Calculating the Amount of an Annuity

An investor deposits $400 every six months for 10 years at 8% per year, compounded semiannually. How much is in the account immediately after the last payment?

Solution

Interest per period: $ i = \frac{0.08}{2} = 0.04 $. Number of payments: $ n = 20 $.

Apply the amount formula: $ A_f = 400\,\frac{(1.04)^{20} - 1}{0.04} $.

$ A_f \approx \$11{,}911.23 $.

EXAMPLE 2 Required Periodic Payment

How much should be invested every month at 12% per year, compounded monthly, to have $4000 in 18 months?

Solution

$ i = \frac{0.12}{12} = 0.01 $, $ A_f = 4000 $, $ n = 18 $.

Solve $ 4000 = R\,\frac{(1.01)^{18} - 1}{0.01} $ for $ R $: $ R = \frac{4000(0.01)}{(1.01)^{18} - 1} \approx 203.93 $.

The monthly investment should be about $203.93.

EXAMPLE 3 Present Value of an Annuity

A lottery pays $500,000 per year for 20 years. If 10% interest is available, compounded annually, what is the present value?

Solution

$ i = 0.1 $, $ R = 500{,}000 $, $ n = 20 $.

$ A_p = 500{,}000\,\frac{1 - (1.1)^{-20}}{0.1} \approx \$4{,}256{,}781.86 $.

EXAMPLE 4 The Amount of a Loan

A student can pay $200/month for 4 years at 12% interest, no down payment. What price car can she afford?

Solution

$ i = \frac{0.12}{12} = 0.01 $, $ R = 200 $, $ n = 48 $.

The price equals the present value of the payments: $ A_p = 200\,\frac{1 - (1.01)^{-48}}{0.01} \approx \$7594.79 $.

EXAMPLE 5 Monthly Mortgage Payments

A couple borrows $100,000 at 9% interest for 30 years (monthly payments). Find each payment.

Solution

$ A_p = 100{,}000 $, $ i = \frac{0.09}{12} = 0.0075 $, $ n = 360 $.

$ R = \frac{(0.0075)(100{,}000)}{1 - (1.0075)^{-360}} \approx \$804.62 $.

Instructor Note

Instructor note — to be added.

SECTION 12.5

Mathematical Induction

— Conjecture and Proof

Checking a finite number of cases can never prove a statement true for all natural numbers, since there are infinitely many. For example, $ p(n) = n^2 - n + 41 $ gives a prime for every $ n $ up to 40 but fails at $ n = 41 $. A proof is a clear argument that establishes a statement's truth beyond doubt.

— Mathematical Induction

Mathematical induction proves a statement $ P(n) $ for all natural numbers using two steps: the base case (prove $ P(1) $ is true) and the induction step (prove that if $ P(k) $ is true then $ P(k+1) $ is true). The assumption that $ P(k) $ is true is called the induction hypothesis. Note that the induction step does not prove $ P(k) $; it only shows the implication.

— Principle of Mathematical Induction

If $ P(1) $ is true, and for every natural number $ k $, $ P(k) $ true implies $ P(k+1) $ true, then $ P(n) $ is true for all natural numbers $ n $.

— Sums of Powers

\[ \sum_{k=1}^{n} 1 = n \] \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \] \[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \] \[ \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} \]

EXAMPLE 1 A Proof by Mathematical Induction

Prove that for all natural numbers $ n $: $ 1 + 3 + 5 + \cdots + (2n-1) = n^2 $.

Solution

Base case: $ P(1) $ says $ 1 = 1^2 $, which is true.

Induction hypothesis: assume $ 1 + 3 + 5 + \cdots + (2k-1) = k^2 $.

Induction step: add the next term $ [2(k+1)-1] $ to both sides: \[ \underbrace{[1 + 3 + \cdots + (2k-1)]}_{= k^2} + [2(k+1)-1] = k^2 + (2k+1) \]

Simplify: $ k^2 + 2k + 1 = (k+1)^2 $, which is $ P(k+1) $.

By the Principle of Mathematical Induction, the statement holds for all $ n $.

EXAMPLE 2 A Proof by Mathematical Induction

Prove that $ 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}{2} $.

Solution

Base case: $ P(1) $: $ 1 = \frac{1(1+1)}{2} = 1 $. True.

Induction hypothesis: assume $ 1 + 2 + \cdots + k = \frac{k(k+1)}{2} $.

Induction step: add $ (k+1) $: \[ \frac{k(k+1)}{2} + (k+1) = (k+1)\left(\frac{k}{2} + 1\right) = (k+1)\cdot\frac{k+2}{2} \]

This equals $ \frac{(k+1)[(k+1)+1]}{2} $, which is $ P(k+1) $.

By induction, the formula holds for all $ n $.

EXAMPLE 3 Proving an Inequality by Induction

Prove that $ 4n < 2^n $ for all $ n \geq 5 $.

Solution

Base case: $ P(5) $: $ 4 \cdot 5 = 20 < 32 = 2^5 $. True.

Induction hypothesis: assume $ 4k < 2^k $ for some $ k \geq 5 $.

Induction step: $ 4(k+1) = 4k + 4 < 2^k + 4 < 2^k + 4k < 2^k + 2^k = 2 \cdot 2^k = 2^{k+1} $.

So $ P(k+1) $ holds, and by induction the inequality is true for all $ n \geq 5 $.

Instructor Note

Instructor note — to be added.

SECTION 12.6

The Binomial Theorem

— Binomial and Its Expansion

An expression of the form $ a + b $ is a binomial. The expansion of $ (a+b)^n $ has $ n+1 $ terms: the first is $ a^n $ and the last is $ b^n $. The exponents of $ a $ decrease by 1 from term to term while the exponents of $ b $ increase by 1, and the sum of the exponents in each term is always $ n $.

— Pascal's Triangle

Pascal's triangle is a triangular array of the coefficients in successive binomial expansions. Key property: every entry (other than a 1) is the sum of the two entries diagonally above it. Each row can be built from the one above, but this recursive method is impractical for large $ n $.

— Factorial and Binomial Coefficient

$ n! $ (n factorial) is the product of the first $ n $ natural numbers, with $ 0! = 1 $ by definition. The binomial coefficient $ \binom{n}{r} $ gives a coefficient in the expansion directly, without building Pascal's triangle.

— Factorial

\[ n! = 1 \cdot 2 \cdot 3 \cdots (n-1)\cdot n, \qquad 0! = 1 \]

— Binomial Coefficient

\[ \binom{n}{r} = \frac{n!}{r!\,(n-r)!}, \qquad 0 \le r \le n \]

— Key Properties of Binomial Coefficients

\[ \binom{n}{r} = \binom{n}{n-r} \] \[ \binom{k}{r-1} + \binom{k}{r} = \binom{k+1}{r} \]

— The Binomial Theorem

\[ (a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n-1}ab^{n-1} + \binom{n}{n}b^n \]

— General Term of the Binomial Expansion

The term containing $ a^r $ in $ (a+b)^n $ is \[ \binom{n}{r}\,a^r b^{\,n-r} \]

EXAMPLE 1 Expanding with Pascal's Triangle

Expand $ (a+b)^7 $ using Pascal's triangle.

Solution

The seventh row of Pascal's triangle is $ 1, 7, 21, 35, 35, 21, 7, 1 $.

Exponents of $ a $ decrease from 7, exponents of $ b $ increase to 7: \[ (a+b)^7 = a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7 \]

EXAMPLE 2 Expanding a Substituted Binomial

Use Pascal's triangle to expand $ (2 + 3x)^5 $.

Solution

Fifth-row coefficients: $ 1, 5, 10, 10, 5, 1 $, so $ (a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 $.

Substitute $ a = 2 $, $ b = 3x $: \[ (2+3x)^5 = 32 + 240x + 720x^2 + 1080x^3 + 810x^4 + 243x^5 \]

EXAMPLE 3 Calculating Binomial Coefficients

Evaluate $ \binom{9}{4} $ and $ \binom{100}{3} $.

Solution

$ \binom{9}{4} = \frac{9!}{4!\,5!} = \frac{6\cdot7\cdot8\cdot9}{1\cdot2\cdot3\cdot4} = 126 $.

$ \binom{100}{3} = \frac{100!}{3!\,97!} = \frac{98\cdot99\cdot100}{1\cdot2\cdot3} = 161{,}700 $.

EXAMPLE 4 Expanding with the Binomial Theorem

Expand $ (x+y)^4 $.

Solution

Coefficients: $ \binom{4}{0}=1,\ \binom{4}{1}=4,\ \binom{4}{2}=6,\ \binom{4}{3}=4,\ \binom{4}{4}=1 $.

\[ (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 \]

EXAMPLE 6 Finding a Particular Term

Find the term containing $ x^5 $ in $ (2x + y)^{20} $.

Solution

Use the general term with $ a = 2x $, $ b = y $, $ n = 20 $, $ r = 5 $: $ \binom{20}{5}(2x)^5 y^{15} $.

$ \binom{20}{5} = \frac{20!}{5!\,15!} $, and $ (2x)^5 = 32x^5 $, giving $ 496{,}128\,x^5 y^{15} $.

EXAMPLE 7 Finding a Coefficient with Negative Powers

Find the coefficient of $ x^8 $ in $ \left(x^2 + \tfrac{1}{x}\right)^{10} $.

Solution

General term: $ \binom{10}{r}(x^2)^r\left(x^{-1}\right)^{10-r} = \binom{10}{r}x^{3r-10} $.

Set the exponent equal to 8: $ 3r - 10 = 8 \Rightarrow r = 6 $.

Coefficient: $ \binom{10}{6} = 210 $.

Instructor Note

Instructor note — to be added.

— Recursive Sequences as Models

Many real-world processes occur in stages: population growth (each generation a new stage), compound interest (each payment a new balance), or measuring a continuously cooling object at fixed intervals. A recursion relation expresses the value at each stage in terms of the value at the preceding stage. In many cases an explicit formula can be found by computing the first few terms and identifying the pattern.

— Compound-Interest Recursion

\[ A_n = (1 + i)\,A_{n-1} \quad\Longrightarrow\quad A_n = (1+i)^n A_0 \]

— Drug-Dose Style Recursion (with constant input)

\[ A_n = c\,A_{n-1} + d \] Solved using the partial sum of a geometric sequence when expanded.

EXAMPLE 1 Population Growth

A population grows 2% each year from an initial 5000. Find a recursive model and an explicit formula for $ P_n $.

Solution

Each year's population is 1.02 times the previous: $ P_n = 1.02\,P_{n-1} $, with $ P_0 = 5000 $.

Computing terms: $ P_1 = (1.02)5000 $, $ P_2 = (1.02)^2 5000 $, and so on.

Explicit formula: $ P_n = (1.02)^n\,5000 $ — a geometric sequence with ratio 1.02.

EXAMPLE 2 Daily Drug Dose

A patient takes a 50-mg pill each morning; the body eliminates 40% of the drug every 24 h. Find a model, an explicit formula, and the long-term amount.

Solution

Each morning 60% remains plus 50 mg: $ A_n = 0.6\,A_{n-1} + 50 $, with $ A_0 = 50 $.

Expanding gives $ A_n = 50(1 + 0.6 + 0.6^2 + \cdots + 0.6^n) $, a geometric partial sum.

\[ A_n = 50\left(\frac{1 - 0.6^{\,n+1}}{1 - 0.6}\right) = 125\left(1 - 0.6^{\,n+1}\right) \]

After 5 days: $ A_5 = 125(1 - 0.6^6) \approx 119 $ mg.

As $ n \to \infty $, $ 0.6^{n+1} \to 0 $, so the amount approaches $ 125 $ mg.

Instructor Note

Instructor note — to be added.

- Figure 12.1 — Graph of a decreasing sequence $ a_n = \tfrac{1}{n} $ (isolated points, not connected). - Figure 12.2 — Fibonacci spiral and its appearance in a nautilus shell. - Figure 12.3 — Cake-halving illustration of the convergent series $ \sum \tfrac{1}{2^k} = 1 $. - Figure 12.4 — Pascal's triangle array of binomial coefficients. - Figure 12.5 — Recursive savings/drug-dose accumulation reaching a steady state.