Tuesday, June 9, 2015

[Pri20150402PTSMAP] A Staircase with Higher Steps


     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.

     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 upside-down (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
Total Skip
# skips
# terms
5 – 3 = 2
2 ¸ 2 = 1
7 – 3 = 4
4 ¸ 2 = 2
9 – 3 = 6
6 ¸ 2 = 3
11 – 3 = 8
8 ¸ 2 = 4
13 – 3 = 10
10 ¸ 2 = 5
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

     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.

     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

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