[S1_20151226NPSW] Finding the General Term of a Sequence (2)

Problem

Introduction

     The above was discussed in this previous article.  The earlier parts of the problem are easy.  The major sticking point is finding the formula for  S_n.  We solved that using factorisation and observation, which I feel is the best way.  But what if you cannot do that and you are desperate (for example, in an exam or test)?

     This article introduces Newton’s Method, which can be used as a back-up method, even though it is not in the regular syllabus.

Solution (Newton’s Method)

 

Remark

     Note that number sequences in “IQ tests” (with no problem contexts) have been debunked.  In our case here, the numbers do have a certain regularity arising from the pattern of dots.  In fact this is an arithmetic progression.  What we are calculating is the sum of an arithmetic progression.  However, Newton’s Method extends beyond arithmetic progressions.

Suitable Levels

* Lower School Mathematics

* GCE ‘O’ Level “Elementary” Mathematics (revision)

* GCE ‘A’ Levels H2 Mathematics (revision)

* other syllabuses that involve number patterns and sequences

* any precocious or independent learner who is interested

[H2_20151123APGP] Factor Theorem with Arithmetic and Geometric Progression

Question

Introduction

     This question tests students on their knowledge of arithmetic and geometric series.  They should also be familiar with Factor Theorem and methods of dealing with polynomials.  Once parts (i) and (ii) are solved, part (iii) is quite straightforward, provided that the student remembers how to deal with surds.

Review of Important Facts

Solution

Heuristics Used

H04. Look for pattern(s)

H05. Work backwards

H10. Simplify the problem

H11. Solve part of the problem

H13* Use Equation / write a Mathematical Sentence

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB HL Mathematics

* other syllabuses that series and Factor Theorem

[Pri20150402PTSMAP] A Staircase with Higher Steps

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 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

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?

Heuristics Used

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

[PriCNPTGN_20150508] Sum of the First Few Natural Numbers

Question

 Introduction

     Numerical patterns are a challenge to learn and to teach.  Those of us who are teachers, usually the better students of our time, tend to think that the pattern is “obvious”, and hence do not bother to explain and/or facilitate classroom discussions regarding patterns.

     This article is about a pattern that involves the sum of the first few natural numbers (or wholenumbers).  This pattern, common in our primary school mathematics patterns, is a special case of the sum of an arithmetic progression, but its formula is not usually taught explicitly.  The story is told of the great mathematician Karl Friedrich Gauss, who figuredout a short-cut for adding up  1 + 2 +... + 100.  Lesser mortals in primary schools are left to struggle with frustration, or to copy “model answers” from their tutors or teachers without understanding how the solutions were obtained.
     As in a previous article, I attempt to illustrate the pattern visually.  I shall highlight the linkages to facilitate discovery of the general method, as well as show how the last part (part(c)) can be solved.

Solution

     Many pupils are able to deduce the answer to part (a) by analysing the differences between successive answers, which is equivalent to asking what must you add to get the next number.  For example, from the 1st number  1,  you add  2  to get the 2nd number  3,  you add  3  to get the third number  6.  To get the 4th number, you add  4  to get  10.  This approach works, but will not help you much for parts (b) and (c) of the question.  The better way is to look for a method that does not require you to keep on adding numbers.  That would allow one to kill all the birds with one stone.  How to do this?

     Imagine the given dots being doubled, rotated and then put together into parellelogram-like matrices.  If you want, you can imagine them as rectangular arrays.  I use the colour orange for the originals, and blue for the copies.

     As you can see, for figure 1, the total number of blue and orange dots is  1 ´ 2.  For figure 2, the total is  2 ´ 3.  For figure 3, the total is  3 ´ 4.  For figure 4, the total is  4 ´ 5.  The pattern is: for whatever number representing the position of the figure, the total is this ordinal number multiplied by another factor that is one more than this number.  Hence the number of original dots (shown in orange) is this product divided by  2.  With this insight, we can fill up the table to answer part (a).  The answer for figure  5  is  30.

We can also figure out that the answer to part (b) is  55,  with the above-mentioned pattern.

For part (c), we can use trial and error or “guess and check”.  Trying  21  gives  21´22 = 462  which does not work.  We try 22:  22´23 = 506.   Yes!  Bingo!

You may also use a calculator to help you.  Since the two unknown factors are close together (they differ only by one), it is almost like multiplying a number with itself, or squaring.  So to guess our number, we may use the square root (the opposite of squaring) to estimate it.  The square root of  506  is about  22.49.  We guess  22  and verify that 22´23 = 506.

Summary

     The pattern involving the sum of the first few whole numbers may be deduced by making a copy of the original figure, rotating it and joining it to form an array.  Just multiply accordingly and then divide by  2  to get the sum for the original figure.  To solve for a figure’s ordinal number (which figure has a certain given number of dots), one may use guess and check, or use square roots.  A primary school pupil should be able to all these without the knowledge of advanced techniques like the sum of arithmetic progressions or quadratic equations.

Heuristics Used

H02. Use a diagram / model

H03. Make a systematic list

H04. Look for pattern(s)

H05. Work backwards

H07. Use guess and check

H09. Restate the problem in another way

Suitable Levels

* Primary School Mathematics (algebra unnecessary)

* GCE ‘O’ Level “Elementary” Mathematics (Number patterns, with algebra)

* GCE ‘A’ Levels H2 Mathematics (Number patterns, with algebra)
* anyone who loves a challenge to unravel a pattern