Mathematical inquiry processes: Test cases and generate examples; conjecture, generalise, and prove. Conceptual field of inquiry: The sum of terms in a sequence, simultaneous equations, algebraic proof, proof by mathematical induction.
The prompt, in giving rise to different lines of inquiry, is suitable for students in years 7 up to 12 (grades 6 to 11). If using the prompt with the younger end of the age range, the teacher might replace the generalisation with a particular case - for example, The sum of the first eight terms of the Fibonacci sequence equals the tenth term subtract one. The students might then create their own generalisation after exploring other cases.
The slides support the following lines of inquiry:
Generating the terms in the Fibonacci sequence;
Summing different terms to find properties of the sequence;
Expressing the terms algebraically and proving properties of the sums are always true;
Proving the generalisation in the prompt and others by mathematical induction; and
Solving problems related to missing terms in Fibonacci-type sequences, including by forming and solving simultaneous equations.
During the question, notice, and wonder phase of the inquiry, the teacher must draw out the students' existing knowledge of the Fibonacci sequence and how it is generated. The teacher's aim is for all students to have a clear understanding before exploring the prompt.
In explaining the sequence, the teacher might refer to Leonardo of Pisa's original problem about the growth of a population of rabbits and the assumptions within his model. Alternatively, students might research the origins of the sequence outside the classroom.
The initial phase of exploration sees students choose different values for n to verify the prompt is true for particular cases.
An interesting line of inquiry that arose in a year 7 class involves testing Fibonacci-type sequences that start with different numbers. For example, the sum of the first five terms of the sequence 1, 4, 5, 9, 14, 23, 37 is 33, which is the seventh term subtract the second one. (In the prompt, you subtract one because that is the second term of the Fibonacci sequence.)
The result can be proved for the first five terms of any Fibonacci-type sequence using algebra (where a and b are the first two terms): a, b, a + b, a + 2b, 2a + 3b, 3a + 5b, 5a + 8b. The sum of the first five terms is 5a + 7b, which is the seventh term subtract b (the second term).
March 2025
Summing terms of the Fibonacci sequence can lead to intriguing results. For example, when you sum the first and fifth terms of any five consecutive terms then divide by three, the result is the middle term of the five. For example, using 5, 8, 13, 21, 34 gives (5 + 34) ÷ 3 = 13. See the slides for more ways to sum terms in the sequence.
To prove the result in the first line of inquiry, students can use an algebraic approach. Any five consecutive terms in a Fibonacci-style sequence can be expressed: a, b, a + b, a + 2b, 2a + 3b. The sum of the first and fifth terms is a + (2a + 3b) = 3a + 3b. Dividing by three gives a + b, which is the middle term.
Year 12 students devised the proofs below for the following statements:
The sum of the first n terms of the Fibonacci sequence is the (n+2)th term subtract one.
The sum of the squares of the first n terms of the Fibonacci sequence equals the product of the nth and (n+1)th terms.
Fibonacci-type sequences involve adding two consecutive terms to get the next one. There are problems to solve in the slides. If the third term is 25 and the sixth term is 107 in a Fibonacci-type sequence, for example, a student could use trial and improvement to find the other terms. Let us say the fourth term is 40. Then the fifth would be 65 and the sixth is 105, which is slightly too low. The fourth term must be greater than 40 so we could try 41.
Another approach to the problem in the fourth line of inquiry is to use algebraic expressions. The first six algebraic terms are a, b, a + b, a + 2b, 2a + 3b, 3a +5b. From these we can form two simultaneous equations for the third and sixth terms: a + b = 25 and 3a +5b = 107. Multiplying the first equation by three gives 3a +3b = 75. The difference between the two equations is now 2b = 32 from which we can deduce that b = 16 and a = 9.