Sequences and Factorials

Sequences Defined by a Recursive Formula

  • Sequences appear in nautilus shells, pinecones, tree branches, and other natural structures.

  • The Fibonacci sequence is a sequence where each term is the sum of the two preceding terms.

    • The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34, …

    • Examples include the Calla Lily (1 petal), Black-Eyed Susan (13 petals), and daisies (21 or 34 petals).

  • Each term depends on the terms that come before it.

  • The Fibonacci sequence is best described using a recursive formula.

    • A recursive formula defines the terms of a sequence using previous terms.

    • It consists of the value of an initial term (or terms) and an equation defining ana_n in terms of preceding terms.

Recursive Formula Example

  • Given:

    • a1=3a_1 = 3

    • an=2an11an = 2a{n-1} - 1, for n2n ≥ 2

  • Finding subsequent terms:

    • a1=3a_1 = 3

Calculation of Terms

  • a2=2a11=2(3)1=5a2 = 2a1 - 1 = 2(3) - 1 = 5

  • a3=2a21=2(5)1=9a3 = 2a2 - 1 = 2(5) - 1 = 9

  • a4=2a31=2(9)1=17a4 = 2a3 - 1 = 2(9) - 1 = 17

  • The first four terms of the sequence are {3, 5, 9, 17}.

Fibonacci Sequence Recursive Formula

  • The recursive formula states the first two terms and defines each successive term as the sum of the preceding two terms.

    • a1=1a_1 = 1

    • a2=1a_2 = 1

    • a<em>n=a</em>n1+an2a<em>n = a</em>{n-1} + a_{n-2}, for n3n ≥ 3

  • To find the tenth term, add the eighth and ninth terms:

    • a<em>10=a</em>9+a8=34+21=55a<em>{10} = a</em>9 + a_8 = 34 + 21 = 55

General Note: Recursive Formula

  • A recursive formula defines each term of a sequence using preceding term(s).

  • Recursive formulas must always state the initial term(s) of the sequence.

Q & A

  • Must the first two terms always be given in a recursive formula?

    • No, many recursive formulas define each term using only one preceding term, requiring only the first term to be defined.

How To: Write the First n Terms of a Sequence Given a Recursive Formula

  1. Identify the initial term, a1a_1, which is given as part of the formula.

  2. To find the second term, a<em>2a<em>2, substitute the initial term into the formula for a</em>n1a</em>{n-1}. Solve.

  3. To find the third term, a3a_3, substitute the second term into the formula. Solve.

  4. Repeat until you have solved for the nthn^{th} term.

Example: Writing the Terms of a Sequence Defined by a Recursive Formula

  • Write the first five terms of the sequence defined by the recursive formula:

    • a1=9a_1 = 9

    • a<em>n=3a</em>n120a<em>n = 3a</em>{n-1} - 20, for n2n ≥ 2

Solution

  • The first term is given.

  • For each subsequent term, replace an1a_{n-1} with the value of the preceding term.

    • n=1,a1=9n = 1, a_1 = 9

    • n=2,a<em>2=3a</em>120=3(9)20=2720=7n = 2, a<em>2 = 3a</em>1 - 20 = 3(9) - 20 = 27 - 20 = 7

    • n=3,a<em>3=3a</em>220=3(7)20=2120=1n = 3, a<em>3 = 3a</em>2 - 20 = 3(7) - 20 = 21 - 20 = 1

    • n=4,a<em>4=3a</em>320=3(1)20=320=17n = 4, a<em>4 = 3a</em>3 - 20 = 3(1) - 20 = 3 - 20 = -17

    • n=5,a<em>5=3a</em>420=3(17)20=5120=71n = 5, a<em>5 = 3a</em>4 - 20 = 3(-17) - 20 = -51 - 20 = -71

  • The first five terms are {9, 7, 1, -17, -71}.

Try It

  • Write the first five terms of the sequence defined by the recursive formula:

    • a1=2a_1 = 2

    • a<em>n=2a</em>n1+1a<em>n = 2a</em>{n-1} + 1, for n2n ≥ 2

  • The first five terms are {2, 5, 11, 23, 47}.

Using Factorial Notation

  • Factorial is a new operator denoted by "!".

  • Factorial asks us to multiply together all the positive integers that come before the number in front of the symbol.

  • Factorial is used heavily in Combinatorics, a branch of mathematics concerned with methods for counting sets.

  • The formulas for some sequences include products of consecutive positive integers.

Definition of n Factorial

  • nn factorial, written as n!n!, is the product of the positive integers from 1 to nn.

    • For example:

      • 4!=4321=244! = 4 \cdot 3 \cdot 2 \cdot 1 = 24

      • 5!=54321=1205! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120

  • An example of a formula containing a factorial is an=(n+1)!a_n = (n+1)!.

    • The sixth term of the sequence can be found by substituting 6 for n:

      • a6=(6+1)!=7!=7654321=5040a_6 = (6+1)! = 7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040

  • The factorial of any whole number nn is n!=n(n1)!n! = n(n-1)!. Therefore, 5!5! can also be thought of as 54!5 \cdot 4!.

A General Note: Factorial

  • nn factorial is a mathematical operation that can be defined using a recursive formula.

  • The factorial of nn, denoted n!n!, is defined for a positive integer nn as:

    • 0!=10! = 1

    • 1!=11! = 1

    • n!=n(n1)(n2)(2)(1)n! = n(n-1)(n-2)…(2)(1) , for n2n ≥ 2

  • The special case 0!0! is defined as 0!=10! = 1.

Q & A

  • Can factorials always be found using a calculator?

    • No, factorials get large very quickly—faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial.

Example: Writing the Terms of a Sequence Using Factorials

  • Write the first five terms of the sequence defined by the explicit formula an=5n(n+2)!a_n = 5n(n+2)!

Solution

  • Substitute n=1,n=2n = 1, n = 2, and so on in the formula.

    • n=1,a1=5(1)(1+2)!=5(3!)=5321=30n = 1, a_1 = 5(1)(1+2)! = 5(3!) = 5 \cdot 3 \cdot 2 \cdot 1 = 30

    • n=2,a2=5(2)(2+2)!=10(4!)=104321=240n = 2, a_2 = 5(2)(2+2)! = 10(4!) = 10 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 240

    • n=3,a3=5(3)(3+2)!=15(5!)=1554321=1800n = 3, a_3 = 5(3)(3+2)! = 15(5!) = 15 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 1800

    • n=4,a4=5(4)(4+2)!=20(6!)=20654321=14400n = 4, a_4 = 5(4)(4+2)! = 20(6!) = 20 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 14400

    • n=5,a5=5(5)(5+2)!=25(7!)=257654321=126000n = 5, a_5 = 5(5)(5+2)! = 25(7!) = 25 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 126000

  • The first five terms are {30, 240, 1800, 14400, 126000}.

Try It

  • Write the first five terms of the sequence defined by the explicit formula an=n(n+1)!2na_n = \frac{n}{(n+1)!2^n}

  • The first five terms are {14\frac{1}{4}, 112\frac{1}{12}, 196\frac{1}{96},12880\frac{1}{2880},18640\frac{1}{8640}}.