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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.