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Habits of Mind Problem #1
The Odd Number Triangle
Due Aug 30, 2011
Create a number triangle by listing consecutive odd numbers as shown below, with each row having one
more number than the preceding one:
Row 1:
Row 2:
Row 3:
Row 4:
Row 5:
1
3 5
7 9 11
13 15 17 19
21 23 25 27 29
a) What is the first number in Row 20? What is the last number in Row 20?
b) Determine the sum of all the numbers in row 1, the sum of all the numbers in row 2,
the sum of all the numbers in row 3, the sum of all the numbers in row 4, the sum of all the
numbers in row 5. What do you think is the sum of the the entries in row 6? Make a table to
express your answers. What do you think is the sum of all the numbers in Row n?
c) What is the mean of the entries in rows 1? What is the mean of the entries in rows 2? What is the
mean of the entries in rows 3? What is the mean of the entries in rows 4? Make a table to
express your answers. What do you conjecture is the mean of Row 20?
d) Compare the row means with the number of each row and the first number in each row. What
pattern do you see? From this pattern, can you guess a formula for the first number in each row?
e) Can you prove that your guess is correct? What does the word “prove” mean in this context?
Adapted from Foundations for Success Mathematics Expectations for the Middle Grades
Further adapted by SRD on Aug-30-2011
Solution:
f) Many students in the class showed the perseverance to write out the number triangle to row 20 to get
the answer. That level effort is admired, but unfortunately, for a couple of people it led to the wrong
answer. Instead, let’s note the pattern for the first element in the row.
Row #
1
Increase
2
3
4
5
6
7
8
9
10
16
17
18
19
20
n
2
4
6
8
10
12
14
16
18
30
32
34
36
38
2(n-1)
1st #
#s in
row
1
3
7
13
21
31
43
57
73
91
241
273
307
343
381
1
2
3
4
5
6
7
8
9
10
16
17
18
19
20
last #
row
sum
1
5
11
19
29
41
55
71
89
109
271
305
341
379
419
1
8
27
64
125
216
n3
row ave
1
4
9
16
25
36
n2
n
Using a pattern of increasing row n by 2(n-1), we get that 381 is the first number and then calculate that
419 is the last number.
g) The sum (by row) for the first 6 rows will be 1, 8, 27, 64, 125 and 216. That is enough data to enable
us to guess that for row n, the sum will be n3.
h) The mean (by row) for the first 6 rows will be 1, 4, 9, 16, 25, 36. Here we note that this might have
been easier if we had figured out part c before part b. The mean for row 20 should be 400.
i)
Students in the class found various ways to describe the 1st number of the nth row. They include:
1) n(n-1) + 1; i.e. the row number times the previous row number plus 1.
2) n2 – (n – 1); i.e. the mean of the row (or square of the row number) minus the previous row #
3) (n – 1)2 + n; i.e. the mean of the previous row plus the row number
j) But how do we prove that this answer is correct? Note: Examples may convince us, but they do not
constitute a proof.
Here is one way we can prove the statement.
Note that the number of odd numbers in a row is equal to the row number. We can use this
information to “count” how many odd numbers occur before the nth row. Since there are (n-1) rows,
each of which has (n-1) elements, the answer would be:
1 + 2 + 3 + … + (n-3) + (n-2) + (n-1)
How many numbers is this?
There are formulas for such things and the answer is n(n-1)/2, but let me show you a way to prove
this is the case. Let’s add all the numbers in these two rows:
1 + 2 + 3 + … + (n-3) + (n-2) + (n-1)
+(n-1)+(n-2)+(n-3)+ … + 3 + 2 + 1
Can you “see” that the answer is n + n + n + … + n + n + n written (n-1) times?
If yes, then we know the answer is n(n-1).
But, this is twice as much as we want because we added each number twice. So the answer is as
previously stated:
n(n-1)/2
But, if there are this many odd numbers before the nth row, we also go past exactly the same number
of even numbers.
What is the n(n-1)/2th even number?
It has to be 2*n(n-1)/2 = n(n-1).
Now the first number in the nth row is clear. It is the next number, n(n-1) + 1.
That is one of the three versions we gave as the answer for part d.