Name: 6.4 Polynomial Functions Polynomial Functions: The expression 3r 2 3r + 1 is a in one variable since it only contains variable, r. KEY CONCEPT Polynomial in One Variable Words A polynomial of degree n in one variable x is an expression of the form, where the coefficients, represent real numbers, a n is not zero, and n represents a nonnegative number. Example 3x 5 + 2x 4 5x 3 + x 2 + 1 n= a 5 = a 4 = a 3 = a 2 = a 1 = a 0 = The degree of a polynomial is the exponent of its variable. The coefficient of the term with the degree. PAGE 331 Polynomial Expression Degree Leading Coefficient Constant 9 Linear x-2 Quadratic 3x 2 + 4x -5 Cubic 4x 3-6 General n a n EXAMPLE 1: Find Degrees and Leading Coefficients State the degree and leading coefficient of each polynomial in one variable. If it is not a polynomial in one variable, explain why. a. 7x 4 + 5x 2 + x 9 b. 8x 2 + 3xy 2y 2
A polynomial equation used to represent a function is called a function. For example, the equation is a quadratic polynomial function, and the equation is a cubic polynomial function. Other polynomial functions can be defined by the general rule. KEY CONCEPT Definition of Polynomial Function Words A polynomial function of a degree n is a continuous function that can be described by an equation of the form where the coefficients represent real numbers, a n is not zero, and n represents a nonnegative integer. Example f(x) =4x 2-3x + 2 n= a 2 = a 1 = a 0 = EXAMPLE 2: Real-World Example Natrue: Refer to the application at the beginning of the lesson (pg. 331). a. Show that the polynomial function f(r) = 3r 2 3r + 1 gives the total number of hexagons when r = 1,2, and 3. Find the values of f(1), f(2), f(3). f(1) = 3r 2 3r + 1 f(2) = 3r 2 3r + 1 f(3) = 3r 2 3r + 1 b. Find the total number of hexagons in a honeycomb with 12 rings. f(12) = 3r 2 3r + 1
EXAMPLE 3: Function Values of Variables Find q(a +1) 2q(a) if q(x) = x 2 + 3x + 4 To evaluate q(a +1), replace x in q(x) with a + 1. And simplify. q(a + 1) = (a + 1) 2 + 3(a + 1) + 4 To evaluate 2q(a), replace x with a in q(x), then multiply the expression by 2. q(x) = x 2 + 3x + 4 2q(a) = 2(a 2 + 3a + 4) Now evaluate q(a+1) 2q(a) q(a+1) 2q(a) = a 2 + 5a + 8 (2a 2 +6a +8) Graphs of Polynomial Functions PG 333 Constant Function: Degree 0 Linear Function: Degree 1 Quadratic Function: Degree2 Cubic Function: Degree 3 Quartic Function: Degree 4 Quintic Function: Degree 5
zeros? How does the degree compare to the maximum number of CONCEPT SUMMARY End Behavior of a Polynomial Function PG 334 Degree: Degree: Degree: Degree: Degree: Leading Leading Leading Leading Leading Coefficient: Coefficient: Coefficient: Coefficient: Coefficient: End Behavior: End Behavior: End Behavior: End Behavior: End Behavior: Domain: Domain: Domain: Domain: Domain: Range: Range: Range: Range: Range: EXAMPLE 4: Graphs of Polynomial Functions For each graph, describe the end behavior determine whether it represents an odd-degree or an even-degree polynomial function, and state the number of real zeros a. b.
PAGE 335 a. b.
Name: 6.5 Analyzing Graphs of Polynomial Functions Graph Polynomial Functions: To graph a polynomial function, make a table of values to find several points and then connect them to make a smooth continuous. Knowing the end behavior of the graph will assist you in completing the sketch of the graph. EXAMPLE 1: Graph a Polynomial Function Graph f(x) = x 4 + x 3 4x 2 4x by making a table of values. x f(x) x f(x) KEY CONCEPT Location Principle Words Suppose y=f(x) represents a polynomial function and a and b are two numbers such that and Then the function has at least one real zero between a and b. Model What is a relative maximum and minimum? What is the a turning point and how can you tell how many you can have in a given polynomial function?
EXAMPLE 2: Locate Zeros of a Function Determine consecutive integer values of x between which each real zero of the function f(x) = x 3 5x 2 + 3x + 2. Then draw the graph. x f(x) Check Your Progress: Determine consecutive integer values of x between which each real zero of the function f(x) = x 3 + 4x 2-6x - 7. Then draw the graph. x f(x) EXAMPLE 3: Graph f(x) = x 3 3x 2 + 5. Estimate the x-coordinates at which the relative
maxima and relative minima occur. x f(x) Look at the table of values and the graph. Check Your Progress: Graph f(x) = x 3 + 4x 2-3. Estimate the x-coordinates at which the relative maxima and relative minima occur.
x f(x) Look at the table of values and the graph. EXAMPLE 4: Real-World Graph a Polynomial Model
Energy: The average fuel (in gallons) consumed by individual vehicles in the Unite States from 1960 to 2000 is modeled by the cubic equation F(t) = 0.025t 3 1.5t 2 + 18.25t + 654, where t is the number of years since 1960. a. Graph the equation 0 5 10 15 20 25 30 35 40 x f(x) b.describe the turning points of the graph and its end behavior. c.what trends in fuel consumption does the graph suggest? Is it reasonable to assume that the trend will continue indefinitely?.
Name: 6.6 Solving Polynomial Equations CONCEPT SUMMARY Factoring Techniques Number of Terms Factoring Technique General Case Any Number Greatest Common Factor two Difference of Two Squares Sum of Two Cubes Difference of Two Cubes three Perfect Square Trinomials General Trinomials Four or more Grouping EXAMPLE 1: GCF Factor 6x 2 y 2 2xy 2 + 6x 3 y Check Your Progress 1A. 18x 3 y 4 + 12x 2 y 3-6xy 2 1B. a 4 b 4 + 3a 3 b 4 + a 2 b 3
EXAMPLE 2: Grouping Factor a 3 4a 2 + 3a 12 Group to find GCF Factor the GCF of each binomial Distributive Property Check Your Progress Grouping 2A. x 2 + 3xy + 2xy 2 + 6y 3 2B. 6a 3 9a 2 b + 4ab 6b 2 EXAMPLE 3: Two or Three Terms Factor each polynomial a. 8x 3 24x 2 + 18x b. m 6 n 6
Check Your Progress Two or Three Terms 3A. 3xy 2 48x 3B. c 3 d 3 + 27 KEY CONCEPT Quadratic Formula An expression that is quadratic in form can be written as au 2 + bu + c for any numbers a,b,c, a 0, where u is some expression in x. The expression au 2 + bu + c is called the form of the original expression. EXAMPLE 4 Write Expressions in Quadratic Form Write each expression in quadratic form, if possible. a. x 4 + 13x 2 + 36 b. 12x 8 x 2 + 10 Check Your Progress Write Expressions in Quadratic Form 4A. 16x 6 625 4B. 9x 10 15x 4 + 9 EXAMPLE 5 GO ON
EXAMPLE 5 Solve Polynomial Equations Solve each equation. a. x 4 13x 2 + 36 = 0 Original Equation Write expression on the left in quadratic form. Factor the Trinomial Factor each difference of squares. Use the Zero Product Property. The solutions are,,,. b. x 3 + 343 = 0 Original Equation This is the sum of two cubes. Sum of two cubes formula with a=x and b=7 Simplify. Zero Product Property The solution of the first equation is. The second equation can be solved b using the Quadratic Formula. The solution of the second equation is.
Check Your Progress Solve Polynomial Equations 5A. x 4 29x 2 + 100 = 0 5B. x 3 + 8 = 0
Name: 6.7 The Remainder and Factor Theorem Synthetic Substitution: Synthetic division can be used to find the value of a function. Consider the polynomial function f(a) = 4a 2 3a + 6. Divide the polynomial by a-2. Synthetic Substitution: Synthetic division can be used to find the value of a function. Consider the polynomial function f(a) = 4a 2 3a + 6. Divide the polynomial by a-2. METHOD 1: Long Division METHOD 2: Synthetic Compare the remainder of 16 to f(2). Replace a with 2. Multiply. Simplify. Notice that the value of f(2) is the same as the remainder when the polynomial is divided by a-2. This illustrates the Theorem. KEY CONCEPT If a polynomial f(x) is divided by x-a, the remainder is the constant f(a), and Remainder Theorem f(x) = q(x) (x a) + f(a) where q(x) is a polynomial with degree one less than the degree of f(x).
When division is used to evaluate a function, it is called synthetic substitution. It is a convenient way of finding the value of a function, especially when the degree of the polynomial is than. EXAMPLE 1 Synthetic Substitution If f(x) = 2x 4 5x 2 + 8x 7, find f(6). METHOD 1 Synthetic Substitution METHOD 2 Direct Substitution If f(x) = 2x 4 5x 2 + 8x 7, find f(6). Replace x with 6. By using direct substitution, f(6) =. Both methods give the results. GO ON
Check Your Progress Synthetic Substitution 1A. If f(x) = 3x 3 6x 2 + x 11, find f(3). METHOD 1 Synthetic Substitution METHOD 2 Direct Substitution If f(x) = 3x 3 6x 2 + x 11, find f(3). Replace x with 3. 1B. If g(x) = 4x 5 + 2x 3 + x 2 1, find f(-1). METHOD 1 Synthetic Substitution METHOD 2 Direct Substitution If g(x) = 4x 5 + 2x 3 + x 2 1, find f(-1). Replace x with -1.
Factors of Polynomials: The synthetic division below shows that the quotient of x 4 + x 3 17x 2 20x + 32 and x-4 is x 3 + 5x 2 + 3x 8. SHOW WORK: Since the remainder is 0, f(4)=0. This means that x 4 is a factor of x 4 + x 3 17x 2 20x + 32. This illustrates the, which is a special case of the Remainder Theorem. KEY CONCEPT Factor Theorem The binomial x a is a factor of the polynomial f(x) if and only if f(a) = 0. EXAMPLE 2: Use the Factor Theorem Show that x + 3 is a factor of x 3 + 6x 2 x - 30. Then find the remaining factors of the polynomial.
Check Your Progress: Use the Factor Theorem 2. Show that x - 2 is a factor of x 3-7x 2 + 4x + 12. Then find the remaining factors of the polynomial. EXAMPLE 3: Find All Factors Geometry: The volume of the rectangular prism is given by V(x) = x 3 + 3x 2 36x + 32. Find the missing measures. The volume of a rectangular prism is x x. You know the measure is, so x 4 is a factor of V(x). Use synthetic division. 4 1 3-36 32 x - 4?? The quotient is. Use this to factor V(x). Volume function Factor Factor trinomial x 2 + 7x - 8 So the missing measures of the prism are and.
Name: 6.8 Roots and Zeros Types of Roots You have learned that a of a function f(x) is any value c such that f(c) =. When the function is graphed, the of the function are the x-intercepts of the graph. Key Concept Page 362 Zeros, Factors, and Roots Let f(x) = be a polynomial function. Then the following statements are equivalent. In addition, if c is a real number, then (, 0) is an of the graph of f(x). Fundamental Theorem of Algebra- all polynomial equations with degree greater than will have at least root in the set of complex numbers. EXAMPLE 1: Determine Number and Type of Roots Solve each equation. State the number and type of roots a. x 2 8x + 16 = 0 Original Equation Factor the left side as a perfect square trinomial.
Solve for x using the Square Root Property Since x-4 is twice a factor of x 2 8x + 16, 4 is a root. So this equation has one real repeated root, 4. b. x 4 1= 0 This equation has real roots, and, and two roots, and. Check Your Progress: Determine Number and Type of Roots 1A. x 3 + 2x = 0 1B. x 4-16 = 0 KEY CONCEPT Corollary
A polynomial equation of the form P(x) of degree n with complex coefficients has exactly n roots in the set of complex numbers. KEY CONCEPT Descartes Rule of Signs pg 363 If P(x) is a polynomial with real coefficients, the terms of which are arranged in descending powers of the variable, EXAMPLE 2 Find Numbers of Positive and Negative Zeros State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x)=x 5 6x 4 3x 3 + 7x 2 8x + 1. Since p(x) has a degree of, it has zeros. However some may be imaginary. Use Descartes rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x). p(x) =x 5 6x 4 3x 3 + 7x 2 8x + 1 to to to to to Since there is sign change, there is exactly negative real zero. Thus the function p(x) has either 4,2, or 0 positive real zeros and exactly 1 negative real zero. Make a chart of the possible combinations of real and imaginary zeros.
# of Positive Real Zeros # of Negative Real Zeros # of Imaginary Zeros Total Number of Zeros EXAMPLE 3: Use Synthetic Substitution to Find Zeros Find all zeros of f(x) = x 3 4x 2 + 6x 4 Since f(x) has a degree of, the function has zeros. To determine the possible number and type of real zeros, examine the number of the sign changes for f(x) and f(-x). f(x) = x 3 4x 2 + 6x 4 to to to f(-x) = -x 3 4x 2-6x 4 to to to Since there are sign changes for the coefficients of f(x), the function has or positive real zeros. Since there are sign changes for the coefficient of f(- x), f(x) has no real zeros. Thus, f(x) has either real zeros, or real zero and imaginary roots. To find these zeros, first list some possibilities and then eliminate those that are not zeros. Since none of the zeros are negative and f(0) is -4, begin by evaluating f(x) for positive integral values from 1 to 4. You can use a shortened form of substitution to find f(a) for several values of a. x -1-4 6-4 1 2
3 4 From the table, we can see that zero occurs at x=. Since depressed polynomial of this zero, x 2-2x+2, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation, x 2-2x+2=0. Quadratic Formula: Thus the function has real zero at x= and imaginary zeros at x= and x=. The graph of the function verifies that there is only. KEY CONCEPT Complex Conjugates Theorem Suppose a and b are real numbers with b 0. If a + bi is a zero of a polynomial function with real coefficients, then a-i is also a zero of the function. EXAMPLE 4 Use Zeros to Write a Polynomial Function Write a polynomial function of least degree with integral coefficients the zeros of which include 3 and 2 i. Explore:
Plan: Multiply the factors to find the polynomial function. Check: 4. Check Your Progress Use Zeros to Write a Polynomial Function Write a polynomial function of least degree with integral coefficients the zeros of which include -1 and 1 + 2i. Explore: Plan: Multiply the factors to find the polynomial function.
Check:
Name: 11.1 Arithmetic Sequence A is a list of numbers in a particular order. Each number in a sequence is called a. The first term is symbolized by a 1, the second term is symbolized as. An sequence is a sequence in which each term after the first is found by adding a constant, called the, to the previous term. EXAMPLE 1 Find the Next Terms Find the next four terms of the arithmetic sequence of 55, 49,43, 55 49 43 1. Check Your Progress Find the next four terms of the arithmetic sequence of -1.6, -0.7,0.2, -1.6-0.7 0.2 Make a Table Sequence Numbers 55 49 43 37 Symbols Expressed in Terms of d Numbers Symbols and the First Term KEY CONCEPT nth Term of an Arithmetic Sequence The nth term a n of an arithmetic sequence with first term a 1 and common difference d is given by the following formula, where n is any positive integer. a n = a 1 + (n-1)d
REAL-WORLD EXAMPLE Find a Particular Term Construction The table to the right shows typical costs for a construction company to rent a crane for one, two, three, or four months. If the sequence continues, how much would it cost to rent the crane for twelve months? Months Cost ($) 1 75,000 2 90,000 Explore: 3 105,000 4 120,000 Plan: Check:
2. Check Your Progress The construction company has a budget of $350,000 for crane rental. The job is expected to last 18 months. Will the company be able to afford the crane rental for the entire job? Explain. Months Cost ($) Explore: 1 75,000 2 90,000 3 105,000 4 120,000 Plan: Check:
EXAMPLE 3 Write an Equation for the nth Term Write n equation for the nth term of the arithmetic sequence 8, 17, 26, 35, In this sequence, a 1 = and d=. Use the nth term formula to write an equation. a n = a 1 + (n-1)d An equation is a n =. 3. Check Your Progress Write an Equation for the nth Term Write n equation for the nth term of the arithmetic sequence -1.5, -3.5, -5.5, In this sequence, a 1 = and d=. Use the nth term formula to write an equation. a n = a 1 + (n-1)d An equation is a n =. The terms between any two nonsuccessive terms of an arithmetic sequence are called. In the sequence below, 41, 52, and 63 are the three arithmetic means between 30 and 74. 19, 30, 41, 52, 63, 74, 85, 96 Three arithmetic means between 30 and 40
EXAMPLE 4 Find the Arithmetic Means You can use the nth term formula to find the common difference. In the sequence 16,,,,, 91,, a 1 is 16 and a 6 is 91. a n = a 1 + (n-1)d Now use the value of d to find the four arithmetic means. 16 91 + + + + + 4. Check Your Progress Find the Arithmetic Means You can use the nth term formula to find the common difference. In the sequence 15.6,,,,, 60.4,, a 1 is 15.6 and a 6 is 60.4. a n = a 1 + (n-1)d Now use the value of d to find the four arithmetic means. 15.6 60.4 + + + + +
Name: 11.2 Arithmetic Series A is an indicated sum of the terms of a sequence. Since 18,22,26,30 is an arithmetic sequence, 18+22+26+30 is an. S n represents the sum of the first n terms of a series. For example, is the sum of the first four terms. To develop a formula for the sum of any arithmetic series, consider the series below. S 9 = 4 + 11 + 18 + 25 + 32 + 39 + 46 + 53 + 60 Write S 9 in two different orders and add the two equations. S 9 = 4 + 11 + 18 + 25 + 32 + 39 + 46 + 53 + 60 + S 9 = An arithmetic series S n has n terms, and the sum of the first and last terms is a 1 + a n. Thus, the formula represents the sum of any arithmetic series. KEY CONCEPT Sum of an Arithmetic Series The sum S n of the first n terms of an arithmetic series is given by S n = [2a 1 + (n-1)d] or S n = (a 1 + a n )
EXAMPLE 1 Find the Sum of an Arithmetic Series Find the sum of the first 100 positive integers. The series 1 + 2 + 3 + + 100. Since you can see that a 1 =, a 100 =, and d=, you can use either sum formula for this series. METHOD 1 METHOD 2 Formula: S n = (a 1 + a n ) Formula: S n = [2a 1 + (n-1)d] 1. Check Your Progress Find the Sum of an Arithmetic Series Find the sum of the first 50 positive integers. The series 1 + 2 + 3 + + 50. Since you can see that a 1 =, a 50 =, and d=, you can use either sum formula for this series. METHOD 1 METHOD 2 Formula: S n = (a 1 + a n ) Formula: S n = [2a 1 + (n-1)d]
EXAMPLE 2 REAL-WORLD Find the First Term Radio A radio station is giving away a total of $124,000 in August. If they increase the amount given away each day by $100, how much should they give away the first day? You know the values of n, S n, and d. Use the sum formula that contains d. Sum Formula that contains d: S n = [2a 1 + (n-1)d] Plug in n, S n, and d: The radio station should give away the first day.
EXAMPLE 3 Find the First Three Terms Find the first three terms of an arithmetic series in which a 1 =9, a n =105, S n =741. Step 1 Since you know a 1, a n, and s n, use S n = (a 1 + a n ) to find n. Step 2 Find d. Step 3 Use d to determine a 2 and a 3. The first three terms are,,.
3. Check Your Progress Find the First Three Terms Find the first three terms of an arithmetic series in which a 1 =-16, a n =33, S n =68. Step 1 Since you know a 1, a n, and s n, use S n = (a 1 + a n ) to find n. Step 2 Find d. Step 3 Use d to determine a 2 and a 3. The first three terms are,,. Writing out a series can be time consuming and lengthy. For convenience, there is a more concise notation called. The series 3 + 6 + 9 + 12 + + 30 can be expressed as. This expression is read The sum of 3n as n goes from 1 to 10. Label the parts of Sigma Notation
The variable, in this case n, is called the index of. EXAMPLE 4 Evaluate a Sum in Sigma Notation Evaluate METHOD 1 Find the terms by replacing j with 5, 6, 7, and 8. Then add. METHOD 2 Since the sum is an arithmetic series, use the formula S n = (a 1 + a n ). There are terms, a 1 = or, and a 4 = or. EXAMPLE 4 Evaluate a Sum in Sigma Notation Evaluate METHOD 1 Find the terms by replacing j with 5, 6, 7, and 8. Then add. METHOD 2 Since the sum is an arithmetic series, use the formula S n = (a 1 + a n ). There are terms, a 1 = or, and a 4 = or.
Name: 11.3 Geometric Sequence The sequence of heights is an example of a. A geometric sequence is a sequence in which each term after the first is found by the previous term by a nonzero constant r called the. Sequence numbers 2 6 18 54 Expressed in Terms of r and the Previous Term Expressed in Terms of r and the First Term symbols a 1 a 2 a 3 a 4 numbers symbols a 1 a 1 r a 2 r a 3 r a n-1 r numbers symbols a 1 r 0 a 1 r 1 a 1 r 2 a 1 r 3 a 1 r n-1 KEY CONCEPT nth Term of a Geometric Sequence The nth term a n of a geometric sequence with the first term a 1 and common ratio r is given by the following formula, where n is any positive integer. a n = a 1 r n-1 EXAMPLE 2 Find the Term Given the First Term and the Ratio Find the eighth term of a geometric sequence for which a 1 = -3 and r = -2. Formula: a n = a 1 r n-1
2. Check Your Progress Find the Term Given the First Term and the Ratio Find the sixth term of a geometric sequence for which a 1 = and r = 3. Formula: a n = a 1 r n-1 EXAMPLE 3 Write an Equation for the nth Term Write an equation for the nth term of the geometric sequence 3, 12, 48, 192, Formula: a n = a 1 r n-1 3. Check Your Progress Write an Equation for the nth Term Write an equation for the nth term of the geometric sequence 18, -3,,, Formula: a n = a 1 r n-1
EXAMPLE 4 Find the Term Given One Term and the Ratio Find the tenth term of a geometric sequence for which a 4 = 108 and r = 3. STEP 1 Find the value of a 1 STEP 2 Plug in a 1 and solve for a 10 Formula: a n = a 1 r n-1 Formula: a n = a 1 r n-1 The tenth term is. 4. Check Your Progress Find the Term Given One Term and the Ratio Find the eighth term of a geometric sequence for which a 3 = 16 and r = 4. STEP 1 Find the value of a 1 STEP 2 Plug in a 1 and solve for a 8 Formula: a n = a 1 r n-1 Formula: a n = a 1 r n-1 The eighth term is.
The missing terms between two nonconsecutive terms of a geometric sequence are called. For example, 6, 18, and 54 are three geometric means between 2 and 162 in the sequence...you can use the common ratio to find the geometric mean in a sequence. EXAMPLE 5 Find the Geometric Means Find three geometric means between 2.25 and 576. Use the nth term formula to find the value of r. In the sequence 2.25,,,, 576, a 1 is 2.25 and a 5 is 576. Formula: a n = a 1 r n-1 There are two possible common ratios, so there are two possible sets of geometric means. Use each value of r to find three geometric means. r = 4 r = -4 The geometric means are,, The geometric means are,,
5. Check Your Progress Find the Geometric Means Find three geometric means between 4 and 13.5. Use the nth term formula to find the value of r. In the sequence 4,,,, 13.5, a 1 is 4 and a 5 is 13.5. Formula: a n = a 1 r n-1 There are two possible common ratios, so there are two possible sets of geometric means. Use each value of r to find three geometric means. r = r = The geometric means are The geometric means are
Name: 11.4 Geometric Series The numbers form a geometric sequence in which a 1 = 1 and r = 3. The indicated sum of the numbers in the sequence is called a geometric series. S 8 = or 3280 The expression for s8 can be written as S 8 =. A rational expression like this can be used to find the sum of any geometric series. KEY CONCEPT The sum S n of the first nth term of a geometric series is given by Sum of Geometric Series Formula: EXAMPLE 1 REAL-WORLD Find the Sum of the First n Terms Health Contagious diseases can spread very quickly. Suppose five people are ill during the first week of an epidemic, and each person who is ill spreads the disease to four people by the end of the next week. By the end of the tenth week of the epidemic, how many people have been affected by the illness? This is a geometric series with a 1 = 5, r = 4 and n = 10. Formula:
EXAMPLE 2 Evaluate a Sum Written in Sigma Notation METHOD 1 Since the sum is a geometric series, you can use the formula Evaluate Formula: METHOD 2 Find the terms by replacing n with 1,2,3,4,5, and 6 and then add. Evaluate Formula: 2. Check Your Progress Evaluate a Sum Written in Sigma Notation METHOD 1 Since the sum is a geometric series, you can use the formula Evaluate Formula: METHOD 2 Find the terms by replacing n with 1,2,3,4,5, and 6 and then add. Evaluate Formula:
How can you find the sum of a geometric series if you know the first and last terms and the common ratio, but not the number of terms? You can use the formula for the nth term of a geometric sequence or series, a n = a 1 r n-1, to find an expression involving r n. Formula: a n = a 1 r n-1 Multiply each side by r: a n r = a 1 r n-1 r a n r = a 1 r n Now substitute a n r for a 1 r n in the formula for the sum of geometric series. The result is S n = EXAMPLE 3 Use the Alternate Formula for a Sum Find the sum of a geometric series for which a 1 = 15,625, a n = -5, and r = Since you do not know the value of n, use the formula derived above. Formula: 3. Check Your Progress Use the Alternate Formula for a Sum Find the sum of a geometric series for which a 1 = 1,000, a n = 125, and r = Since you do not know the value of n, use the formula derived above. Formula:
EXAMPLE 4 Find the First Term of a Series Find a 1 in a geometric series for which S 8 = 39,360 and r = 3. Formula: 4. Check Your Progress Find the First Term of a Series Find a 1 in a geometric series for which S 7 = 258 and r = -2. Formula: