BINOMIAL_THEOREM_2009

					               South Western Sydney Region HSC Mathematics Extension Study Day
                University of Western Sydney Campbelltown, 22nd September 2009
                           THE BINOMIAL THEOREM
                      Robert Yen, Hurlstone Agricultural High School
Outline
  1. Introduction                                       6. Finding the greatest coefficient
  2. Binomial expansions and Pascal’s triangle          7. Proving identities involving the sum of
     n                                                                  n
  3. Ck, a formula for Pascal’s triangle                   coefficients Ck
  4. The binomial theorem in the past 10 HSC exams      8. Binomial probability
  5. Finding a particular term                          9. How to study for Maths: a 4-step approach


1. INTRODUCTION
      The PowerPoint presentation that accompanies these notes can be found at the HSC Online
       website: http://hsc.csu.edu.au/maths
      This topic examines the general pattern for expanding (a + x)n
      It is a difficult topic because it involves new work on high-level algebra and is learned at the
       end of the course with little time for practice and revision
      HSC questions involving this topic are often targeted at better Extension 1 students,
       especially when they appear in Question 7, so if you are aiming to achieve at the highest
       band (E4) in this course, work on mastering this topic to excel in the exam
      There are no shortcuts to success in this topic: you just have to learn the theory to develop a
       full understanding

2. BINOMIAL EXPANSIONS AND PASCAL’S TRIANGLE
                       Binomial expansion                 No. of terms Coefficients of terms
       (a + x)1 = a + x                                        2                1 1
              2    2          2
       (a + x) = a + 2ax + x                                   3              1 2 1
       (a + x)3 = a3 + 3a2x + 3ax2 + x3                        4             1 3 3 1
       (a + x)4 = a4 + 4a3x + 6a2x2 + 4ax3 + x4                5           1 4 6 4 1
       (a + x)5 = a5 + 5a4x + 10a3x2 + 10a2x3 + 5ax4 + x5      6        1 5 10 10 5 1

The expansion of (a + x)n has n + 1 terms, with the powers of a decreasing from n to 0 and the
powers of x increasing from 0 to n. The sum of the powers in each term is always n. The
coefficients of the terms appear in Pascal’s triangle, where each number is the sum of the two
numbers above it.

3. nCk, A FORMULA FOR PASCAL’S TRIANGLE
      n
        Ck from the Permutations and combinations topic also gives the value of row n, term k of
       Pascal’s triangle, if the top of the triangle is row 0 and the first term in each row is term 0
                                                                         0
              1                                                        C0
                                                                     1
            1 1                                                       C0 1C1
                                                                   2
           1 2 1                                                    C0 2C1 2C2
                                                                 3
          1 3 3 1                                                 C0 3C1 3C2 3C3
                                                               4
         1 4 6 4 1                                              C0 4C1 4C2 4C3 4C4
                                                             5
        1 5 10 10 5 1                                         C0 5C1 5C2 5C3 5C4 5C5
                                                           6
       1 6 15 20 15 6 1                                     C0 6C1 6C2 6C3 6C4 6C5 6C6
                                                         7
     1 7 21 35 35 21 7 1                                  C0 7C1 7C2 7C3 7C4 7C5 7C6 7C7
                                                       8
    1 8 28 56 70 56 28 8 1                              C0 8C1 8C2 8C3 8C4 8C5 8C6 8C7 8C8
                                                                                   n
      C stands for coefficient as well as combination, and nCk is also written as  
                                                                                   k 
                                                                                    
                                       5
       There are 3 ways of calculating C3:
                            5 4 3 5 4
       (a) Mentally: 5C3 =                   = 10
                            3  2 1     2
                            5!     120                         n!
       (b) Formula: 5C3 =       =      = 10, using nCk =
                           3!2! 6  2                      k!(n  k )!
                                       5  4  3 5  4  3 2 1        5!
           This works for 5C3 because                                  .
                                        3  2  1 3  2  1 2  1 3!  2!
       (c) Calculator key: pressing 5 nCr 3 = gives 10.


The binomial theorem

   (a + x)n = nC0 an + nC1 an-1 x + nC2 an-2 x2 + nC3 an-3 x3 + nC4 an-4 x4 + … + nCn xn

    or in sigma notation:
                                                                           n
                   n                                                           Ck is the coefficient
                                   nk k
   (a + x) =n
                       n
                            Ck a      x                                         in the term that
                                                                               contains xk in the
                 k 0
                                                                                    expansion

    the sum of terms                the general term
    from k = 0 to n

Properties of nCk
1. nC0 = nCn = 1                          First and last coefficients are 1
2. nC1 = nCn-1 = n                        Second and second-last coefficients are n
3. nCk = nCn-k                            Pascal’s triangle is symmetrical, for example, 6C2 = 6C4
4. n+1Ck = nCk-1 + nCk                    Pascal’s triangle result: each coefficient is the sum of the two
                                          coefficients in the row above it

Example 1
   Use the binomial theorem to expand:                               Answers

       (a) (a + 3)5                                                   a5 + 15a4 + 90a3 + 270a2 + 405a + 243

       (b) (2x – y)4                                                 16x4 – 32x3y + 24x2y2 – 8xy3 + y4




                                      The binomial theorem: Robert Yen (page 2)
4. THE BINOMIAL THEOREM IN THE PAST 10 HSC EXAMS
         HSC          Finding a           Finding the         Proving              Binomial
         exam       particular term        greatest          identities           probability
                                          coefficient
         1999                            last tested in        Q7(b)                 Q3(b)
         2000           Q2(b)            1988, Q6(b)
         2001           Q2(d)                                  Q5(b)                Q5(c)
         2002                                                  Q7(b)                Q4(a)
         2003           Q2(d)                                                 Q3(c), Q8(a) Ext 2
         2004                                                  Q7(b)                Q4(c)
         2005           Q2(b)                                                       Q6(a)
         2006                                                 Q2(b)                 Q6(b)
         2007                                              Q6(a)(i) Ext 2           Q4(a)
         2008      Q1(d), Q6(c)(i)                           Q6(c)(ii)


5. FINDING A PARTICULAR TERM
Example 2 (2008 HSC, Question 1(d), 2 marks)
   Find an expression for the coefficient of x8y4 in the expansion of (2x + 3y)12.
                                                                              [Answer: 12C4 (28)(34)]


Steps for finding a particular term

1. Write a formula for the general term Tk of the expansion and simplify the formula,
       for example, Tk = 12Ck (2x)12-k (3y)k.

2. To find the term with the required power of x, solve an equation for k,
        for example, 12 – k = 8, or k = 4.
    k must be a whole number or you have made a mistake.

3. Substitute the value of k into the Tk formula to find the required term.


Tk is not the kth term
     In the expansion of (a + x)n, Tk is the term that contains xk
     It is not the kth term but actually the (k + 1)th term, for example, T3 is the 4th term (T0, T1, T2,
       T3), the one that contains x3
     It is simpler to write out the first few terms of the expansion rather than try to memorise the
       sigma notation
     It is also better to avoid referring to the ‘kth term’ and calling its formula Tk+1 (as some
       textbooks do) because students can get confused about the value of k to substitute (in
       Example 2 above, some substituted k = 5 ‘for the 5th term’ instead of k = 4)
     Anyway, HSC questions tend to ask you to find, for example, ‘the term that contains x8’
       rather than ‘the 9th term’

Example 3 (2005 HSC, Question 2(b), 3 marks)
                                                                                                     12
                                                                                          1 
    Use the binomial theorem to find the term independent of x in the expansion of  2 x  2  .
                                                                                         x 
                                                                                [Answer: 126 720]


                                The binomial theorem: Robert Yen (page 3)
Common student mistakes
   Giving the position of the term (‘the 5th term’) rather than the actual term
   Poor use of algebra, index laws, brackets and negative signs
   Wasting time expanding out all the terms
   Substituting wrong value for k, such as k + 1 instead
   In Example 2, giving the coefficient as 12C4 only instead of 12C4 (28)(34)


6. FINDING THE GREATEST COEFFICIENT
     In (1 + 2x)8 = 1 + 16x + 112x2 + 448x3 + 1120x4 + 1792x5 + 1792x6 + 1024x7 + 256x8, the
greatest coefficient is 1792 (occurring twice).
     The term with the greatest coefficient usually occurs in the middle of an expansion because
with the rows in Pascal’s triangle, the larger numbers are in the middle. In any expansion of
(a + x)n, the coefficients usually increase, reach a maximum, then decrease.

Example 4
                             8
    Suppose (1 + 2x)8 =     t x
                            k 0
                                    k
                                            k
                                                .

    (a) Find an expression for tk, the coefficient of xk.                                 [Answer: 8Ck 2k]
                   t     2 8  k 
    (b) Show that k 1             .
                     tk    k 1
    (c) Show that the greatest coefficient is 1792.


Steps for finding the greatest coefficient

1. Write formulas for the general coefficient tk and the next coefficient tk+1.
            t                                   b( n  k )
2. Simplify k 1 to an expression of the form              .
              tk                                a  k  1

    Note that
                n  1! = n and        c n 1
                                               = c1 = c.
                                           n
                   n                     c
         t k 1
3. Solve        > 1 to find the highest integer value of k.
          tk
4. Find the value of tk+1, the greatest coefficient.


HOMEWORK EXERCISE (1988 HSC, Question 6(b), 6 marks)
                              25
    Suppose (7 + 3x)25 =     t x
                             k 0
                                        k
                                                k
                                                    .

    (i) Use the binomial theorem to write an expression for tk, 0  k  25.
                    t     3(25  k )
    (ii) Show that k 1             .
                      tk   7(k  1)
    (iii) Hence or otherwise find the largest coefficient tk.
                                                   25 
          You may leave your answer in the form  7 c 3 d .
                                                  k
                                                   
                                                                                     25 
                                                                     [Answer: t 7   71837 (≈ 1.71 × 10 24 )]
                                                                                    7
                                                                                     



                                        The binomial theorem: Robert Yen (page 4)
7. PROVING IDENTITIES INVOLVING THE SUM OF COEFFICIENTS nCk

                                       Pascal’s triangle                        n         ΣnCk                    Σ(nCk)2
                                                                                     sum of values           sum of (values)2
                                         1                                      0        1 (20)                   1 (1C0)
                                        11                                      1        2 (21)                   2 (2C1)
                                       121                                      2        4 (22)                   6 (4C2)
                                      1331                                      3        8 (23)                  20 (6C3)
                                     14641                                      4       16 (24)                  70 (8C4)
                                   1 5 10 10 5 1                                5       32 (25)                 252 (10C5)
                                 1 6 15 20 15 6 1                               6       64 (26)                 924 (12C6)
                                1 7 21 35 35 21 7 1                             7      128 (27)                3432 (14C7)
                              1 8 28 56 70 56 28 8 1                            8      256 (28)               12 870 (16C8)

Two important identities:
      n
1.   
     k 0
            n
                Ck  2 n
                                  5
     For example,            k 0
                                      5
                                          Ck  5C0  5C1  5C2  5C3  5C4  5C5  1  5  10  10  5  1  32  25 .
                       2

                  Ck  
      n
                n            2n
2.                                Cn
     k 0


                                         Ck    4C0    4C1    4C2    4C3    4C4  .
                                  4                    2            2       2          2                 2    2
                                           4
     For example,
                               k 0

                                                            12  42  62  42  12  70  8C4

Identities involving the sum of coefficients can be proved by expanding (1  x)n and then:
    substituting x = 0, 1 or -1, or
    equating coefficients, or
    differentiating or integrating.


The binomial theorem for (1 + x)n

       (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + … + nCn xn
                                                                                      n
       or in sigma notation:                                            (1 + x)n =   
                                                                                     k 0
                                                                                            n
                                                                                                Ck x k



Example 5
                                                                                                                   n
       Expand (1 + x)n and substitute an appropriate value of x to prove that                                     
                                                                                                                  k 0
                                                                                                                         n
                                                                                                                             Ck  2 n .


Example 6
   By considering that (1 + x)2n = (1 + x)n(1 + x)n and examining the coefficient of xn on each

                              C 
                              n
                                                   2
side, prove that                       n
                                               k        2 nC n .
                             k 0




                                                             The binomial theorem: Robert Yen (page 5)
Hints for proving identities (by John Dillon, Head Teacher of Maths, Hurlstone AHS)
                              (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + …+ nCn xn
If the identity involves ...                                  try ...
n
  Ck’s with no x’s                                            substituting a simple value such as x = 0 or x = 1
n
  Ck’s with alternating + and – signs                         substituting a negative value for x
powers of a number (say a) as well as nCk’s                   substituting x = a
n
  Ck’s multiplied by k’s                                      differentiating both sides
n
  Ck’s divided by (k + 1)’s                                   integrating both sides but don’t forget ‘+ c’

HOMEWORK EXERCISES
1. Expand both sides of the identity (1 + x)n(1 + x) = (1 + x)n+1 and compare coefficients to prove
   Pascal’s triangle result n+1Ck = nCk-1 + nCk.
                                                                               n
2. Expand (1 + x)n and differentiate both sides to prove that                k  C
                                                                              k 1
                                                                                     n
                                                                                         k    n  2n 1 .


Example 7 (2006 HSC, Question 2(b), 2 marks)
   (i) By applying the binomial theorem to (1 + x)n and differentiating, show that                                           1
                                      n n                 n                 n
                     n 1  x      2   x  ...  r   x r 1  ...  n   x n 1.
                               n 1

                                      1   2               r                  n
   (ii) Hence deduce that                                                                                                    1
                                        n           n                  n
                               n3n 1     ...  r   2r 1  ...  n   2n 1.
                                        1           r                 n

Example 8 (2008 HSC, Question 6(c), 5 marks)
   Let p and q be positive integers with p  q.

    (i) Use the binomial theorem to expand (1 + x)p + q, and hence write down the term                                       2
             1  x 
                        pq
                                                                                                                   pq 
        of                     which is independent of x.                                [Answer to (i) and (ii):      ]
                 xq                                                                                                q 
                              1  x 
                                         pq                    q
                                            1
                                 1  x  1   , apply the binomial theorem and the result of
                                                       p
    (ii) Given that                                                                                                          3
                                            x
                            q
                          x
                                                       p  q   p  q     p  q 
        part (i) to find a simpler expression for 1 +             .
                                                       1  1   2  2     p  p 

Common student mistakes
   Messy and careless working, unclear notation; not enough working, ‘fudging’ the answer
   Forgetting the first term nC0 or the last term nCn xn
   Using series formulas or mathematical induction instead of the binomial theorem: this
    usually doesn’t work
   Not realising that the parts of the question are related
   Getting lost in sigma notation; from the examiners’ notes on the 2008 HSC exam (p.6):
             ‘Responses that used sigma notation were sometimes less successful than (students) who wrote
             out the sum showing at least three correct terms. Many ... misinterpreted this part of the
             question by stating which term was independent of x rather than by giving the independent
             term or, by being careless in their notation, failed to gain this mark.’
       If integrating, forgetting the constant at the end
       Starting a proof using the identity to be proved, rather than prove that LHS = RHS


                                               The binomial theorem: Robert Yen (page 6)
Example 9 (2002 HSC, Question 7(b), 6 marks, HARD!)
   The coefficient of xk in (1 + x)n, where n is a positive integer, is denoted by ck (so ck = nCk).

    (i) Show that c0 + 2c1 + 3c2 + … + (n + 1)cn = (n + 2) 2n-1.                                         3
                       c      c     c                   cn
    (ii) Find the sum 0  1  2  ...  (1) n                   .                                       3
                      1 2 2  3 3  4            (n  1)(n  2)
                                                                                                  1
       Write your answer as a simple expression in terms of n.                        [Answer:       ]
                                                                                                 n2

8. BINOMIAL PROBABILITY (for you to study at home)
     With binomial probability, we are concerned with repeated trials in which there are only two
possible outcomes: we can call one outcome a success, with probability p, and the other outcome a
failure, with probability q = 1 – p. Examples of such outcomes are heads vs. tails, win vs. lose, true
vs. false, boy vs. girl, defective vs. working.


             If a binomial trial is repeated n times, then the probability of r successes is
                                       P(X = r) = nCr pr qn-r
                  X is called the random variable and its value ranges from 0 to n.


Example 10 (2007 HSC, Question 4(a), 4 marks)
   In a large city, 10% of the population has green eyes.

    (i) What is the probability that two randomly chosen people both have green eyes?            1
                                                                                     [Answer: 0.01]
    (ii) What is the probability that exactly two of a group of 20 randomly chosen people        1
          have green eyes? Give your answer correct to three decimal places.
                                                                                    [Answer: 0.285]
    (iii) What is the probability that more than two of a group of 20 randomly chosen people     2
          have green eyes? Give your answer correct to two decimal places.
                                                                                     [Answer: 0.32]

Note that this is an application of the binomial theorem, because:
     P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + ... + P(X = 20)
      = 20C0 0.10 0.920 + 20C1 0.11 0.919 + 20C2 0.12 0.918 + 20C3 0.13 0.917 + …+ 20C20 0.1200.90
      = (0.9 + 0.1)20 that is, (q + p)n                                              Each probability is a
      = 120                                                                         term of the expansion
      = 1.                                 The sum of the                                of (q + p)n
                                         probabilities of all
                                         possible events is 1
Common student mistakes
    Not using the complementary result as a shortcut
    Forgetting to include P(X = 0)
    Does ‘more than two’ include two?




                               The binomial theorem: Robert Yen (page 7)
HOMEWORK EXERCISE (2004 HSC, Question 4(c))
    Katie is one of ten members of a social club. Each week one member is selected at random to
win a prize.

    (i) What is the probability that in the first 7 weeks Katie will win at least 1 prize?        1
                                                                                             7
                                                                                    9
                                                                       [Answer: 1 –    0.5217 ]
                                                                                     10 
    (ii) Show that in the first 20 weeks Katie has a greater chance of winning exactly 2 prizes  2
         than of winning exactly 1 prize.
                                                    [Answer: P(X = 2) ≈ 0.2852 > P(X = 1) ≈ 0.2702]
    (iii) For how many weeks must Katie participate in the prize drawing so that she has         2
         a greater chance of winning exactly 3 prizes than of winning exactly 2 prizes?
                                                                                 [Answer: 30 weeks]


9. HOW TO STUDY FOR MATHS: A 4-STEP APPROACH (P-R-A-C)
1. PRACTISE YOUR MATHS
     Master your skills, strengthen your ability
     Achieve a high level of understanding

2. REWRITE YOUR MATHS
     Summarise the theory and examples in your own words
     Work through all topics to see the big picture
     Achieve an overview of the whole course

3. ATTACK YOUR MATHS
     Identify your areas of weakness and work on overcoming them
     Fill in any gaps in your mathematical knowledge

4. CHECK YOUR MATHS
     Revise your understanding on mixed revision exercises and past HSC exams

Before an exam
   Review and memorise your topic summaries
   Practise on your weak areas
   Practise on HSC-style questions
   Anticipate the exam: the format and structure, the style of questions, planning your time
    during the exam.

Useful resources
   The NSW HSC Online website has tips, tutorials and links: http://hsc.csu.edu.au/maths
   The Board of Studies website has past HSC exams and Notes from the Marking Centre
    (examiners’ reports): www.boardofstudies.nsw.edu.au/hsc_exams
   The Mathematical Association of NSW sells booklets of past HSC exams with worked
    solutions; you may be able to buy these through your school: www.mansw.nsw.edu.au




                               The binomial theorem: Robert Yen (page 8)

				
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