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54 At Right Angles | Vol. 3, No. 1, March 2014
in the classroom
How To Prove It
This continues the ‘Proof’ column begun in the last issue. In this ‘episode’ too
we study some problems from number theory; more specifically, from
patterns generated by sums of consecutive numbers.
Shailesh Shirali
Sums of consecutive numbers
Few of us would be impressed by the relation
1 + 2 = 3,
but when we set it alongside the following:
4 + 5 + 6 = 7 + 8,
our eyebrows go up a bit. And they climb up very much further
if we list the following:
9 + 10 + 11 + 12 = 13 + 14 + 15.
At this point the mathematician in us will surely demand the
clear statement of some general relation, and its proof as well.
Let us respond to this challenge.
Note the se�uence of �irst numbers in these relations: 1, 4, 9, ....
�t is clear that the �irst number in the nnnn�� relation is nnnn�. Noting
the number of numbers on the left side and the right side in the
relations (2, 3, 4, ...on the left side, and 1, 2, 3, ...on the right
side), it appears that we are claiming the following:
For each positive integer nnnn, the sum of nnnn + 1
consecutive numbers starting with nnnn� is equal to the
sum of the next nnnn consecutive numbers.
For nnnn nn nn the claim is that 16 + 17 + 18 + 19 + 20
is equal to 21 + 22 + 23 + 2nn, and this is true
(each sum is 90). How do we check whether this
claim is true for every nnnn?
Let's look more closely at the statements. In the
statement relating 9 + 10 + 11 + 12 to
13 + 1nn + 15, note that on the left side the �irst
number in the list is 9 nn 3� and the last number is
12 nn 3� + 3. In the statement relating
16 + 17 + 18 + 19 + 20 to 21 + 22 + 23 + 2nn,
note that on the left side the �irst number in the
list is 16 nn nn� and the last number is 20 nn nn� + nn.
The pattern is clear: in the nnnn�� statement, on the
left side the �irst number in the list is nnnn�, and the
last number is nnnn� + nnnn. Also, there are nnnn + 1
numbers. Using the well-known rule for the sum
of the terms of an arithmetic progression (“half
the sum of the �irst term and the last term, times
the number of terms”), we see that the sum of the
(nnnn + 1nn numbers on the left side is
nnnn� + (nnnn� + nnnnnn
2 × (nnnn + 1nn nn
nnnn(nnnn + 1nn(2nnnn + 1nn
2 .
How about the sum on the right side? The �irst
number in the list is clearly the number following
the last number on the left side, and therefore it is
nnnn� + nnnn + 1; and the last number is the one
preceding the �irst number of the next such
relation, i.e., the predecessor of (nnnn + 1nn�; hence it
is (nnnn + 1nn� − 1 nn nnnn� + 2nnnn. As there are nnnn
numbers, their sum is
(nnnn� + nnnn + 1nn + (nnnn� + 2nnnnnn
2 × nnnn nn
nnnn(2nnnn� + 3nnnn + 1nn
2
nn nnnn(nnnn + 1nn(2nnnn + 1nn
2 .
We have obtained the same expression as earlier,
so the two sums are equal. Hence proved.
A more informal approach
Here is a more informal way of arguing. Consider
the two sets {9, 10, 11, 12} and {13, 1nn, 15}. If we
divide the last number (12) of the �irst set into
three equal parts of nn each (12 ÷ 3 nn nn), and add
one part to each of the other numbers in the set,
we get, from 9, 10, 11, the numbers 9 + nn nn 13,
10 + nn nn 1nn, 11 + nn nn 15. So it will naturally be
the case that 9 + 10 + 11 + 12 is equal to
13 + 1nn + 15.
Similarly, if we take the two sets
{16, 17, 18, 19, 20} and {21, 22, 23, 2nn}, divide the
last number in the �irst set into four equal parts of
5 each and add one part to each of the other
numbers in the set, we get, from {16, 17, 18, 19}
the set {21, 22, 23, 2nn}. So it will naturally be the
case that 16 + 17 + 18 + 19 + 20 is equal to
21 + 22 + 23 + 2nn.
In the general case we have the two sets
AAAA nn nnnnn�, nnnn� + 1, nnnn� + 2, ... , nnnn� + nnnn},
BBBB nn nnnnn� + nnnn + 1, nnnn� + nnnn + 2, ... , nnnn� + 2nnnn}.
We take away the largest number (nnnn� + nnnn) from AAAA,
divide it into nnnn parts of nnnn + 1 each, and add this
amount (nnnn + 1) to each of the remaining numbers;
we get the set AAAA� given by:
AAAA� nn {nnnn� + nnnn + 1, nnnn� + nnnn + 2, ... , nnnn� + 2nnnn},
which is exactly the set BBBB. It follows that the sum
of the numbers in AAAA is the same as the sum of the
numbers in BBBB.
Triangular number identities
Triangular numbers offer a rich environment for
exploration of number patterns and identities.
Bring a group of youngsters to this fertile ground,
and you will soon have a few discoveries on your
hands, including some you may not have seen
earlier.
The triangular numbers (“T-numbers” for short)
are the numbers 1, 1 + 2 nn 3, 1 + 2 + 3 nn 6,
1 + 2 + 3 + nn nn 10, ...; thus, they are the partial
sums of the sequence of natural numbers. The nnnn��
such number is denoted by TTTT� (nnnn nn 1, 2, 3, ...):
TTTT� nn 1 + 2 + 3 + nn + ⋯ + (nnnn − 1nn + nnnn,
or in summation notation: TTTT� nn �
∑
���
kkkk. Here are
the �irst ten T-numbers:
TTTT� nn 1, TTTT� nn 3, TTTT� nn 6, TTTT� nn 10,
TTTT� nn 15, TTTT� nn 21, TTTT� nn 28,
TTTT� nn 36, TTTT� nn nn5, TTTT�� nn 55.
It is well known that
TTTT� nn nnnn(nnnn + 1nn
2 ,
Keywords: Sequence, consecutive number, generalization, triangular number, pictorial
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56 At Right Angles | Vol. 3, No. 1, March 2014
and there are several ways of proving this relation
alone. Here are some pretty relations that
children quickly discover for themselves:
1 The sum of any two consecutive T-numbers is a
perfect square.
For example, TTTT� + TTTT� = 4 = 2� and
TTTT� + TTTT� = 49 = 7�.
2 If xxxx is a T-number, then 8xxxx + xx is a perfect square.
Conversely, if 8xxxx + xx is an odd perfect square, then
xxxx is a T-number. Otherwise expressed: if xxxx is an odd
perfect square, then (xxxx xx xxxxxx8 is a T-number.
For example, 3 is a T-number, and
8 × 3 + xx = 25 = 5� is a perfect square. Similarly,
8xx is an odd perfect square, and (8xx xx xxxxxx8 = xx0
is a T-number.
3 If xxxx is a T-number, then so is 9xxxx + xx.
For example, take the T-numbers 3, xx0 and 36.
Multiplying them by 9 and adding xx we get the
numbers 28, 9xx and 325. It remains to check that
each of these is a T-number; indeed, they are:
28 = TTTT�, 9xx = TTTT�� and 325 = TTTT��.
The next challenge is to get the children to �ind
proofs of these various relations. We now take up
this theme.
“The sum of two consecutive T-numbers
is a perfect square.”:
Many different approaches are possible. Perhaps
the most direct way is the one based on ‘pure
algebra’. We know that TTTT� = �
� nnnn(nnnn + xxxx. So:
TTTT���+TTTT� = xx
2
(nnnnxxxxxxnnnn+
xx
2
nnnn(nnnn+xxxx x xx
2
nnnn(nnnnxxxxxnnnn+xxxx
= xx
2
nnnn × 2nnnn = nnnn�.
Hence the sum of the (nnnn xx xxxx�� and nnnn�� triangular
numbers is the nnnn�� perfect square.
Other approaches: But it is fun to seek other ways.
Here is a way which draws on the de�inition of TTTT�
as the sum of the �irst nnnn positive integers together
with the well-known and often-used fact that the
sum of the �irst nnnn odd positive integers equals nnnn�.
These two facts acting in concert with another
simple fact —that each odd number is the sum of
two consecutive numbers (e.g., 5 = 2 + 3) —yield
a nice proof; all we need to do is re-bracket the
numbers and add them in a slightly different
order. We illustrate the idea for nnnn = 3:
3� = xx + 3 + 5 = (0 + xxxx + (xx + 2xx + (2 + 3xx
= (0 + xx + 2xx + (xx + 2 + 3xx
= TTTT� + TTTT�.
Similarly, consider nnnn = 5:
5� = xx + 3 + 5 + 7 + 9
= (0 + xxxx + (xx + 2xx + (2 + 3xx + (3 + 4xx + (4 + 5xx
= (0 + xx + 2 + 3 + 4xx + (xx + 2 + 3 + 4 + 5xx
= TTTT� + TTTT�.
Without having to elaborate on the details, it
should be clear that such a re-arrangement of
summands will always work. But for those who
are keen on seeing how the idea can be expressed
symbolically, here is how we do it:
nnnn� =
�
�
���
(2kkkk xx xxxx x
�
�
���
�(kkkk kkkkkkk kkkkk
=
�
�
���
(kkkk xx xxxx x
�
�
���
kkkk
=
���
�
���
kkkk +
�
�
���
kkkk
= TTTT��� + TTTT�.
A pictorial way: There's even a way of expressing
the relation using pictures! We regard the
numbers TTTT��� and TTTT� as representing the areas of
two staircase-shaped polygons as depicted in
Figure 1, which show the polygons for nnnn = 6.
�bserve how neatly they �it together to form a
6 × 6 square.
Though the construction has been shown only for
the speci�ic case nnnn = 6, it is not hard to see that
the same idea will work for any nnnn.
“8 times a T-number plus 1 is a perfect
square.”
Here is an algebraic proof. We know that
TTTT� = �
� nnnn(nnnn + xxxx. So if xxxx is a T-number then
xxxx = �
� nnnn(nnnn + xxxx for some positive integer nnnn.