Given that F1002 1.138 * 10209, find an approximate value for F1000 in scientific

Compute the value of each of the following. (a) F15 (b) F15 - 2 (c) F15-2 (d) F15 5 (e) F15>5

Compute the value of each of the following. (a) F16 (b) F16 + 1 (c) F16+1 (d) F16 4 (e) F16>4

Compute the value of each of the following. (a) F1 + F2 + F3 + F4 + F5 (b) F1+2+3+4+5 (c) F3 * F4 (d) F3*4

Compute the value of each of the following. (a) F1 + F3 + F5 + F7 (b) F1+3+5+7 (c) F10>F5 (d) F10>F5

Describe in words what each of the expressions represents. (a) 3FN + 1 (b) 3FN+1 (c) F3N + 1 (d) F3N+1

Describe in words what each of the expressions represents. (a) F2N - 3 (b) F2N-3 (c) 2FN - 3 (d) 2FN-3

Given that F36 = 14,930,352 and F37 = 24,157,817, (a) find F38. (b) find F39

Given that F32 = 2,178,309 and F33 = 3,524,578, (a) find F34. (b) find F35

Given that F36 = 14,930,352 and F37 = 24,157,817, (a) find F35. (b) find F34.

Given that F32 = 2,178,309 and F33 = 3,524,578, (a) find F31. (b) find F30.

Using a good calculator (an online calculator if necessary) and Binets simplified formula, compute F20.

Using a good calculator (an online calculator if necessary) and Binets simplified formula, compute F25.

Consider the following sequence of equations involving Fibonacci numbers. 1 + 2 = 3 1 + 2 + 5 = 8 1 + 2 + 5 + 13 = 21 1 + 2 + 5 + 13 + 34 = 55 f (a) Write down a reasonable choice for the fifth equation in this sequence. (b) Find the subscript that will make the following equation true. F1 + F3 + F5 + g+ F21 = F? (c) Find the subscript that will make the following equation true (assume N is odd). F1 + F3 + F5 + g+ FN = F?

Consider the following sequence of equations involving Fibonacci numbers. 2(2) - 3 = 1 2(3) - 5 = 1 2(5) - 8 = 2 2(8) - 13 = 3 f (a) Write down a reasonable choice for the fifth equation in this sequence. (b) Find the subscript that will make the following equation true. 21F?2 - F15 = F12 (c) Find the subscript that will make the following equation true. 21FN+22 - FN+3 = F?

Fact: If we make a list of any four consecutive Fibonacci numbers, the first one times the fourth one is always equal to the third one squared minus the second one squared. (a) Verify this fact for the list F8, F9, F10, F11. (b) Using the list FN, FN+1, FN+2, FN+3, write this fact as a mathematical formula

Fact: If we make a list of any 10 consecutive Fibonacci numbers, the sum of all these numbers divided by 11 is always equal to the seventh number on the list. (a) Verify this fact for the list F1, F2, . . . , F10. (b) Using the list FN, FN+1, . . . , FN+9, write this fact as a mathematical formula.

Express each of the following as a single Fibonacci number. (a) FN+1 + FN+2 = (b) FN - FN-2 = (c) FN + FN+1 + FN+3 + FN+5 =

Express each of the following as a single Fibonacci number. (a) FN-2 + FN-3 = (b) FN+2 - FN = (c) FN-3 + FN-2 + FN + FN+2 =

Express each of the following as a ratio of two Fibonacci numbers. (a) 1 + FN FN-1 = (b) FN-1 FN - 1 =

Express each of the following as a ratio of two Fibonacci numbers. (a) 1 + FN-1 FN = (b) 1 - FN FN-2 =

Exercises 21 through 24 refer to Fibonacci-like sequences. Fibonacci-like sequences are based on the same recursive rule as the Fibonacci sequence (from the third term on each term is the sum of the two preceding terms), but they are different in how they get started.Consider the Fibonacci-like sequence 5, 5, 10, 15, 25, 40, . . . , and let AN denote the Nth term of the sequence. (a) Find A10. (b) Given that F25 = 75,025, find A25. (c) Express AN in terms of FN

Exercises 21 through 24 refer to Fibonacci-like sequences. Fibonacci-like sequences are based on the same recursive rule as the Fibonacci sequence (from the third term on each term is the sum of the two preceding terms), but they are different in how they get started.Consider the Fibonacci-like sequence 2, 4, 6, 10, 16, 26, . . . , and let BN denote the Nth term of the sequence. (a) Find B9. (b) Given that F20 = 6765, find B19. (c) Express BN in terms of FN+1.

Exercises 21 through 24 refer to Fibonacci-like sequences. Fibonacci-like sequences are based on the same recursive rule as the Fibonacci sequence (from the third term on each term is the sum of the two preceding terms), but they are different in how they get started.Consider the Fibonacci-like sequence 1, 3, 4, 7, 11, 18, 29, 47, . . . , and let LN denote the Nth term of the sequence. (Note: This sequence is called the Lucas sequence, and the terms of the sequence are called the Lucas numbers.) (a) Find L12. (b) The Lucas numbers are related to the Fibonacci numbers by the formula LN = 2FN+1 - FN. Verify that this formula is true for N = 1, 2, 3, and 4. (c) Given that F20 = 6765 and F21 = 10,946, find L20.

Exercises 21 through 24 refer to Fibonacci-like sequences. Fibonacci-like sequences are based on the same recursive rule as the Fibonacci sequence (from the third term on each term is the sum of the two preceding terms), but they are different in how they get started.Consider the Fibonacci-like sequence 1, 4, 5, 9, 14, 23, 37, . . . , and let TN denote the Nth term of the sequence. (a) Find T12. (b) The numbers in this sequence are related to the Fibonacci numbers by the formula TN = 3FN+1 - 2FN. Verify that this formula is true for N = 1, 2, 3, and 4. (c) Given that F20 = 6765 and F21 = 10,946, find T20.

Exercises 25 through 29 involve solving quadratic equations using the quadratic formula. Here is an instant refresher on the quadratic formula (for a more in-depth review, any high school algebra book should do): n To use the quadratic formula the quadratic equation must be in the standard form ax2 + bx + c = 0. If the equation is not in standard form, you need to get it into that form. n The solutions of the quadratic equation ax2 + bx + c = 0 are given by x = 1 -b { 2b2 - 4ac2 >2a. The formula gives two different solutions unless b2 - 4ac = 0.Consider the quadratic equation x2 = x + 1. (a) Use the quadratic formula to find the two solutions of the equation. Give the value of each solution rounded to five decimal places. (b) Find the sum of the two solutions in (a). (c) Explain why the decimal part has to be exactly the same in both solutions.

Exercises 25 through 29 involve solving quadratic equations using the quadratic formula. Here is an instant refresher on the quadratic formula (for a more in-depth review, any high school algebra book should do): n To use the quadratic formula the quadratic equation must be in the standard form ax2 + bx + c = 0. If the equation is not in standard form, you need to get it into that form. n The solutions of the quadratic equation ax2 + bx + c = 0 are given by x = 1 -b { 2b2 - 4ac2 >2a. The formula gives two different solutions unless b2 - 4ac = 0.Consider the quadratic equation x2 = 3x + 1. (a) Use the quadratic formula to find the two solutions of the equation. Give the value of each solution rounded to five decimal places. (b) Find the sum of the two solutions in (a). (c) Explain why the decimal part has to be exactly the same in both solutions.

Exercises 25 through 29 involve solving quadratic equations using the quadratic formula. Here is an instant refresher on the quadratic formula (for a more in-depth review, any high school algebra book should do): n To use the quadratic formula the quadratic equation must be in the standard form ax2 + bx + c = 0. If the equation is not in standard form, you need to get it into that form. n The solutions of the quadratic equation ax2 + bx + c = 0 are given by x = 1 -b { 2b2 - 4ac2 >2a. The formula gives two different solutions unless b2 - 4ac = 0.Consider the quadratic equation 3x2 = 8x + 5. (a) Use the quadratic formula to find the two solutions of the equation. Give the value of each solution rounded to five decimal places. (b) Find the sum of the two solutions found in (a).

Exercises 25 through 29 involve solving quadratic equations using the quadratic formula. Here is an instant refresher on the quadratic formula (for a more in-depth review, any high school algebra book should do): n To use the quadratic formula the quadratic equation must be in the standard form ax2 + bx + c = 0. If the equation is not in standard form, you need to get it into that form. n The solutions of the quadratic equation ax2 + bx + c = 0 are given by x = 1 -b { 2b2 - 4ac2 >2a. The formula gives two different solutions unless b2 - 4ac = 0.Consider the quadratic equation 8x2 = 5x + 2. (a) Use the quadratic formula to find the two solutions of the equation. Give the value of each solution rounded to five decimal places. (b) Find the sum of the two solutions found in (a).

Exercises 25 through 29 involve solving quadratic equations using the quadratic formula. Here is an instant refresher on the quadratic formula (for a more in-depth review, any high school algebra book should do): n To use the quadratic formula the quadratic equation must be in the standard form ax2 + bx + c = 0. If the equation is not in standard form, you need to get it into that form. n The solutions of the quadratic equation ax2 + bx + c = 0 are given by x = 1 -b { 2b2 - 4ac2 >2a. The formula gives two different solutions unless b2 - 4ac = 0.Consider the quadratic equation 55x2 = 34x + 21. (a) Without using the quadratic formula, show that x = 1 is one of the two solutions of the equation. (b) Without using the quadratic formula, find the second solution of the equation. (Hint: The sum of the two solutions of ax2 + bx + c = 0 is given by -b>a.)

Consider the quadratic equation 89x2 = 55x + 34. (a) Without using the quadratic formula, show that x = 1 is one of the two solutions of the equation. (b) Without using the quadratic formula, find the second solution of the equation. (Hint: The sum of the two solutions of ax2 + bx + c = 0 is given by -b>a.)

Consider the quadratic equation 21x2 = 34x + 55. (a) Without using the quadratic formula, show that x = -1 is one of the two solutions of the equation. (b) Without using the quadratic formula, find the second solution of the equation. (Hint: The sum of the two solutions of ax2 + bx + c = 0 is given by -b>a.)

Consider the quadratic equation 34x2 = 55x + 89. (a) Without using the quadratic formula, show that x = -1 is one of the two solutions of the equation. (b) Without using the quadratic formula, find the second solution of the equation. (Hint: The sum of the two solutions of ax2 + bx + c = 0 is given by -b>a.)

Consider the quadratic equation 1FN2x2 = 1FN-12x + FN-2, where FN-2, FN-1, and FN are consecutive Fibonacci numbers. (a) Show that x = 1 is one of the two solutions of the equation. [Hint: Try Exercises 29(a) or 30(a) first.] (b) Find the second solution of the equation expressed in terms of Fibonacci numbers. [Hint: Try Exercises 29(b) or 30(b) first.]

Consider the quadratic equation 1FN-22x2 = 1FN-12x + FN, where FN-2, FN-1, and FN are consecutive Fibonacci numbers. (a) Show that x = -1 is one of the two solutions of the equation. [Hint: Try Exercises 31(a) or 32(a) first.] (b) Find the second solution of the equation expressed in terms of Fibonacci numbers. [Hint: Try Exercises 31(b) or 32(b) first.]

The number 1 f is the reciprocal of the golden ratio. (a) Using a calculator, compute 1 f to 10 decimal places. (b) Explain why 1 f has exactly the same decimal part as f. 1Hint: Show that 1 f = f - 1.2

The square of the golden ratio is the irrational number f2 = 11 + 25 2 2 2 = 3 + 25 2 . (a) Using a calculator, compute f2 to 10 decimal places. (b) Explain why f2 has exactly the same decimal part as f.

Given that F499  8.6168 * 10103, find an approximate value for F500 in scientific notation. (Hint: FN>FN-1  f.)

Given that F1002  1.138 * 10209, find an approximate value for F1000 in scientific notation. (Hint: FN>FN-1  f.)

The Fibonacci sequence of order 2 is the sequence of numbers 1, 2, 5, 12, 29, 70, . . . . Each term in this sequence (from the third term on) equals two times the term before it plus the term two places before it; in other words, AN = 2AN-1 + AN-2 1N 32. (a) Compute A7. (b) Use your calculator to compute to five decimal places the ratio A7>A6. (c) Use your calculator to compute to five decimal places the ratio A11>A10. (d) Guess the value (to five decimal places) of the ratio AN>AN-1 when N 7 11.

The Fibonacci sequence of order 3 is the sequence of numbers 1, 3, 10, 33, 109, . . . . Each term in this sequence (from the third term on) equals three times the term before it plus the term two places before it; in other words, AN = 3AN-1 + AN-2 1N 32. (a) Compute A6. (b) Use your calculator to compute to five decimal places the ratio A6>A5. (c) Guess the value (to five decimal places) of the ratio AN>AN-1 when N 7 6.

R and R are similar rectangles. Suppose that the width of R is a and the width of R is 3a. (a) If the perimeter of R is 41.5 in., what is the perimeter of R? (b) If the area of R is 105 sq. in., what is the area of R?

O and Oare similar O-rings. The inner radius of O is 5 ft, and the inner radius of Ois 15 ft. (a) If the circumference of the outer circle of O is 14p ft, what is the circumference of the outer circle of O? (b) Suppose that it takes 1.5 gallons of paint to paint the O-ring O. If the paint is used at the same rate, how much paint is needed to paint the O-ring O?

Triangles T and T shown in Fig. 23 are similar triangles. (Note that the triangles are not drawn to scale.) FiGuRE 23 5 in. 60 m T T (a) If the perimeter of T is 13 in., what is the perimeter of T (in meters)? (b) If the area of T is 20 sq. in., what is the area of T (in square meters)?

Polygons P and P shown in Fig. 24 are similar polygons. FiGuRE 24 2 5 P P (a) If the perimeter of P is 10, what is the perimeter of P? (b) If the area of P is 30, what is the area of P?

Find the value of c so that the shaded rectangle in Fig. 25 is a gnomon to the white 3 by 9 rectangle. (Figure is not drawn to scale.) FiGuRE 25

Find the value of x so that the shaded figure in Fig. 26 is a gnomon to the white rectangle. (Figure is not drawn to scale.)

Find the value of x so that the shaded figure in Fig. 27 is a gnomon to the white rectangle. (Figure is not drawn to scale.) FiGuRE 27

Find the value of x so that the shaded figure in Fig. 28 is a gnomon to the white rectangle. (Figure is not drawn to scale.) FiGuRE 28 x 1 2 6

Rectangle A is 10 by 20. Rectangle B is a gnomon to rectangle A. What are the dimensions of rectangle B?

Find the value of x so that the shaded frame in Fig. 29 is a gnomon to the white x by 8 rectangle. (Figure is not drawn to scale.) FiGuRE 29 1 1

In Fig. 30 triangle BCA is a 36-36-108 triangle with sides of length f and 1. Suppose that triangle ACD is a gnomon to triangle BCA. (a) Find the measure of the angles of triangle ACD. (b) Find the length of the three sides of triangle ACD. FiGuRE 30 1 A C B D 36 108 36 f

Find the values of x and y so that in Fig. 31 the shaded triangle is a gnomon to the white triangle ABC. FiGuRE 31 8 4 6 x y

Find the values of x and y so that in Fig. 32 the shaded figure is a gnomon to the white triangle. FiGuRE 32 6 3 5 4 x y

Find the values of x and y so that in Fig. 33 the shaded triangle is a gnomon to the white triangle. FiGuRE 33 9 15 12 x y

Consider the sequence of ratios FN FN+1 . (a) Using a calculator compute the first 14 terms of this sequence in decimal form (rounded to six decimal places when needed). (b) Explain why 1 FN FN+1 2 S f - 1 1i.e., as N gets larger and larger, the ratios FN FN+1 get closer and closer to f - 12.

Consider the sequence of ratios FN+2 FN . (a) Using a calculator compute the first 15 terms of this sequence in decimal form (rounded to six decimal places when needed). (b) Explain why 1 FN+2 FN 2 S f + 1 1i.e., as N gets larger and larger, the ratios FN+2 FN get closer and closer to f + 12.

Consider the sequence T given by the following recursive definition: TN+1 = 1 + 1 TN , and T1 = 1. (a) Find the first six terms of the sequence, and leave the terms in fractional form. (b) Explain why TN S f (i.e., as N gets larger and larger, TN gets closer and closer to f2.

Consider the sequence U given by the following recursive definition: UN+1 = 1 1 + UN , and U1 = 1. As N gets larger and larger, the terms of this sequence get closer and closer to some number. Give the number expressed in terms of the golden ratio f. (Hint: Try Exercises 55 and 57 first.)

Lucas numbers. The Lucas sequence is the Fibonacci-like sequence 1, 3, 4, 7, 11, 18, 29, 47, . . . (first introduced in Exercise 23). The numbers in the Lucas sequence are called the Lucas numbers, and we will use LN to denote the Nth Lucas number. The Lucas numbers satisfy the recursive rule LN = LN-1 + LN-2 ( just like the Fibonacci numbers), but start with the initial values L1 = 1, L2 = 3. (a) Show that the Lucas numbers are related to the Fibonacci numbers by the formula LN = 2FN+1 - FN. [Hint: Let KN = 2FN+1 - FN, and show that the numbers KN satisfy exactly the same definition as the Lucas numbers (same initial values and same recursive rule).] (b) Show that 1 LN+1 LN 2 S f. 3Hint: Use (a) combined with the fact that 1 FN+1 FN 2 S f.4

a) Explain what happens to the values of 11 - 25 2 2N as N gets larger. (Hint: Get a calculator and experiment with N = 6, 7, 8, . . . until you get the picture.) (b) Explain why FN S fN 25 . [Hint: Use (a) and the original Binets formula.] (c) Using (b), explain why 1 FN+1 FN 2 S f.

Explain why the only even Fibonacci numbers are those having a subscript that is a multiple of 3.

Show that FN+1 2 - FN2 = 1FN-12 1FN+22.

Explain why the shaded figure in Fig. 34 cannot have a square gnomon.

Find the values of x and y so that in Fig. 35 the shaded triangle is a gnomon to the white triangle. (Figure is not drawn to scale.) FiGuRE 35 x y 156 144 60

Let ABCD be an arbitrary rectangle as shown in Fig. 36. Let AE be perpendicular to the diagonal BD and EF perpendicular to AB as shown. Show that the rectangle BCEF is a gnomon to the rectangle ADEF. FiGuRE 36 ED C FA

In Fig. 37 triangle BCD is a 72-72-36 triangle with base of length 1 and longer side of length x. (Using this choice of values, the ratio of the longer side to the shorter side is x>1 = x.) FiGuRE 37 x 36 1 72 A B C D

Show that each of the diagonals of the regular pentagon shown in Fig. 38 has length f.

(a) A regular decagon (10 sides) is inscribed in a circle of radius 1. Find the perimeter in terms of f. (b) Repeat (a) with radius r. Find the perimeter in terms of f and r.

Generic Fibonacci-like numbers. A generic Fibonacci-like sequence has the form a, b, b + a, 2b + a, 3b + 2a, 5b + 3a, . . . (i.e., the sequence starts with two arbitrary numbers a and b and after that each term of the sequence is the sum of the two previous terms). Let GN denote the Nth term of this sequence. (a) Show that generic Fibonacci-like numbers are related to the Fibonacci numbers by the formula GN = bFN-1 + aFN-2. [Hint: Try Exercise 59(a) first.] (b) Show that 1GN+1>GN2 S f. [Hint: Try Exercise 59(b) first.]

You are designing a straight path 2 ft wide using rectangular paving stones with dimensions 1 ft by 2 ft. How many different designs are possible for a path of length (a) 4 ft? (b) 8 ft? (c) N ft? (Hint: Give the answer in terms of Fibonacci numbers.)

Show that F1 + F2 + F3 + g+ FN = FN+2 - 1.

Show that F1 + F3 + F5 + g+ FN = FN+1. (Note that on the left side of the equation we are adding the Fibonacci numbers with odd subscripts up to N.)

Show that every positive integer greater than 2 can be written as the sum of distinct Fibonacci numbers.

In Fig. 39, ABCD is a square and the three triangles I, II, and III have equal areas. Show that x>y = f.

During the time of the Greeks the star pentagram shown in Fig. 40 was a symbol of the Brotherhood of Pythagoras. Consider the three segments of lengths x, y, and z shown in the figure.