Question
Introduction
This pertains to the
sum of consecutive numbers with constant skips. I set this question
to illustrate the heuristic of looking for patterns
[H04]. It is
similar to
this question, except that now the numbers jump or skip by 2
instead of just 1. The more knowledgeable reader will doubtless
recognise this to be an
arithmetic
progression. The challenge now is
how can a primary school pupil do it without having learnt about any more
advanced mathematics or algebra, relying purely on
pattern recognition.
Solution
As in the previous solution, imagine the sum as a series of
vertical bars. The numbers all jump
by 2
this time. Because the jump
amount 2
is constant, you see a nice staircase pattern (shown in violet). Each step of the staircase is of height 2 units. If we
make a copy of it and turn it upsidedown (shown in green), the two staircases
join together nicely to form a rectangle.
Notice that 101+3 = 99+5 = 97+7 =
... etc and they are all equal to 104. If we know the number of columns, we can work
out our desired sum. How many columns
are there?
The number of columns is the same as the number of terms in our sum.
OK, but then how many terms are there?
How to calculate this? Let us
look at a few simple cases first [H10.
Simplify the problem].
Let us try to observe the
pattern. Note that the size of each skip
is always 2. If there are
2 terms, it is just 3
and 5, there is one skip of 2.
From 3 to
7, there are 3
terms, there are two skips of
2 each. From
3 to 9,
there are 4 terms,
the difference is 6 and there are
3 skips. From
3 to 11,
there are 5 terms,
the difference is 8 and there are
4 skips. If you go from 3
to 13, the net jump is 10 and
there are 5 skips
and 6 terms.
We can tabulate the data into a table [H02] below:
skip
size = 2
Start

End

Total Skip

# skips

# terms

3

5

5 – 3 = 2

2 ¸ 2 = 1

2

3

7

7 – 3 = 4

4 ¸ 2 = 2

3

3

9

9 – 3 = 6

6 ¸ 2 = 3

4

3

11

11 – 3 = 8

8 ¸ 2 = 4

5

3

13

13 – 3 = 10

10 ¸ 2 = 5

6

Do you notice some things? [H04]
The total skip is the difference
between the starting and ending numbers.

The number of skips is the
difference divided by the skip size.

The number of terms is always one
more than the number of skips.

Since our last term is 103,
the total skip is 101 – 3 =
98. The number of skips is 98 ¸ 2 = 49. So there are
50 terms i.e. 50
columns.
Hence the size of our rectangle is 50 × 104.
But we only want half of this rectangle (shown in violet). Hence the sum is ½ × 50 × 104 = 2 600.
Ans: 3 + 5 + 7 + ... + 99 + 101 = 2 600
Summary
This article
illustrates the heuristic
[H04 Look for pattern(s)]. Our first pattern we notice is the staircase
pattern. After making a copy and turning
that around, we notice that it forms a rectangle, with columns of size 104
each. Now we look for a pattern
that enables us to find the number of columns, which is the number of terms in
our sum. We note that the number of
terms is always the same as the number of skips, which is the same as the
difference between the start and the end all divided by the skip size. This enables us to solve the challenge in a
way similar to my
previous example.
Reflections
Do you think this method will work for different starting numbers and different ending numbers? For different skip sizes? Why not set up your own similar question and try it yourself and see whether it works?
H02. Use a diagram / model
H04. Look for pattern(s)
H09. Restate the problem in
another way
H10. Simplify the problem
Suitable Levels
* Primary School Mathematics
* GCE ‘O’ Level “Elementary” Mathematics (Number patterns,
with algebra)
* GCE ‘A’ Levels H2 Mathematics (sequences and series, with
algebra)
* IB Mathematics (sequences and series, with algebra)
* anyone who loves patterns and relishes a challenge