Fibonachi

Lesson Plan

Each student writes down two numbers vertically, with the smallest number first. For example, a student might write this :

Each student adds his or her two numbers. For example, the student who wrote 3 and 8 would now have 11:

Student continue adding the previous two numbers until they have ten numbers in a column, including their original two numbers. For example, the student who wrote 3 and 8 would have the following numbers:

128

207

335

Student write the sums of their ten numbers, cover their answers, and raise their hands. The teacher goes around the room giving each student his or her sum, which greatly impresses them! The sum for our example is 869. (My students knew me well enough to know that I did not come close to having that kind of mathematical quickness! They knew that a shortcut or trick must be involved.)

Explain the shortcut. Students multiply the seventh term by 11, which will give them the sum of their ten numbers. In our examples the seventh term is 79, and 11 x 79 = 869, the sum of our ten numbers. (Again, my students were surprised at how quickly I could multiply by 11!)

Explain the trick for multiplying quickly by 11. This trick is best illustrated by showing some examples :

11 x 34 = 374

4 is the ones digit in the number 34

7 is the sum of the ones digit (4) and the tens digit (3) in the number 34

3__ is the tens digit in the number 34

374

11 x 45

5 is the ones digit in 45

9 is the sum of 4 and 5

4 is the tens digit in 45

495

Students’ understanding is enhanced if they multiply the number using the traditional method. For the second example above, they get the folowing :

x 11

45__

495

Once students practice the shortcut with several simpler problems, many of them discover that they can actually use the same trick when working with larger numbers. For example, the following:

11 x 23426

8 (2 + 6)

6 (4 + 2)

7 (3 + 4)

5 (2 + 3)

2______

257,686

Sometimes, adding two of the numbers requires students to regroup when using this shortcut trick. The following example illustrates that small challenge:

11 x 57

2 (5 + 7 = 12, but we regroup the 1 to the next place)

6 (the 5 plus the 1 that was regrouped).

627

Notice that the first examples used did not require regrouping. This procedure is introduced once the students are comfortable with the shortcut for multiplying by 11.

Student look at one another’s sequences, giving sums by using these new strategies.
Finally, demonstrate the proof of why 11 times the sevent term will give the sums of students’ number :

Let x represent the firs number and y, the second number. Remember that each additional number is the sum of previous two numbers.

x First number

y Second number

x + y Third number

x + 2y Fourth number

2x + 3y Fifth number

3x + 5y Sixth number

5x + 8y Seventh number

8x + 13y Eighth number

13x + 21y Ninth number

21x + 34y Tenth number

55x + 88y Sum of the ten numbers.

In the final calculation, 55x and 88y = 11 (5x +8y). Note that (5x + 8y) represents the seventh number. The result will work with any two starting numbers, even fractions and negative numbers.

Extentions

· This approach will work for any two starting numbers, but what if the larger number happens to be listed first? What is the sum of the ten digits that start as 5, -3? How is that problem different from –3,5 ?

· If the third number in such a Fibonacci like sequnce is 11 and the sixth number is 49, what are the first two number? Reffering to the proof in question 7, we can find the answer by using simultaneous equations. The equation for the third number is x + y = 11 and for the sixth number is 3x + 5y = 49 ( Note: This problem can also be solved by trial and error, pairing 2 and 9, 3 and 8, 4 and 5, and so on to give younger children the opportunity to solve the problem or as a second way for algebra students to solve the problem.)

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