Unit 12.1 Patterns and Algebra Topic 9: Binomial Theorem Continuing the focus on different methods of solving real-life problems, Topic 9 looks at the Binomial Theorem. In particular, and in line with the Syllabus (p. 18), this Topic is about: • Discussing and applying Pascal’s Triangle in expanding binomials. • Discussing and applying Binomial Theorem and its properties. Binomial expansions A binomial has two terms. Expansion of binomials raised to powers, of the form (a + b) n where n ∈ whole numbers, gives the following pattern: n = 0, (a + b) 0 = 1 n = 1, (a + b) 1 = a + b n = 2, (a + b) 2 = a 2 + 2ab + b 2 n = 3, (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 n = 4, (a + b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 It is clear from the pattern above that the powers of a and b ascend or descend by 1 from n, and that their sum is n for each term in the expansion. For example, when n = 4 the first term has a 4 (and b 0 ), the second term has a 3 (and b 1 ), the third term has a 2 (and b 2 ), and so on. A method of evaluating the coefficients of the terms was established by Blaise Pascal (1623–1662), who arranged the coefficients in the shape of a triangle. Pascal’s Triangle 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 and so on. Each number in this pattern, other than 1, is the sum of the two numbers standing above it to the left and to the right, eg 15 = 10 + 5. For example, using the last line written above in Pascal’s Triangle allows expansion of (a + b) 7 . (a + b) 7 = a 7 + 7a 6 b + 21a 5 b 2 + 35a 4 b 3 + 35a 3 b 4 + 21a 2 b 5 + 7ab 6 + b 7 Using this array of coefficients allows expansion of (a + b) n , where n is a whole number. SAMPLE PAGES
9
Embed
Discussing and applying Binomial Theorem and its …...Binomial Theorem. In particular, and in line with the Syllabus (p. 18), this Topic is about: • Discussing and applying Pascal’s
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Unit 12.1 Patterns and AlgebraTopic 9: Binomial Theorem
Continuing the focus on different methods of solving real-life problems, Topic 9 looks at the Binomial Theorem. In particular, and in line with the Syllabus (p. 18), this Topic is about:• Discussing and applying Pascal’s Triangle in expanding binomials.• Discussing and applying Binomial Theorem and its properties.
Binomial expansionsA binomial has two terms. Expansion of binomials raised to powers, of the form (a + b)n where n ∈ whole numbers, gives the following pattern:
n = 0, (a + b)0 = 1
n = 1, (a + b)1 = a + b
n = 2, (a + b)2 = a2 + 2ab + b2
n = 3, (a + b)3 = a3 + 3a2b + 3ab2 + b3
n = 4, (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
It is clear from the pattern above that the powers of a and b ascend or descend by 1 from n, and that their sum is n for each term in the expansion. For example, when n = 4 the first term has a4 (and b0), the second term has a3 (and b1), the third term has a2 (and b2), and so on.
A method of evaluating the coefficients of the terms was established by Blaise Pascal (1623–1662), who arranged the coefficients in the shape of a triangle.
Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
and so on.
Each number in this pattern, other than 1, is the sum of the two numbers standing above it to the left and to the right, eg 15 = 10 + 5.
For example, using the last line written above in Pascal’s Triangle allows expansion of (a + b)7.
Unit 12.1 Activity 9A: ExpandingUsing Pascal’s Triangle, answer the following questions:1. Expand the following: a. (1 + x)5 b. (1 – a)8 c. (x + y)5 d. (x – y)6
e. (1 – 3y)6 f. (2 + 3x)5 g. (4 – x)4 h. (1 – 12
x)5
2. Find the first four terms in the expansion of: a. (1 – 2x)9 b. (x + 2y)7 c. (x – 2y)10
3. a. Expand (a + b)4
b. Use your expansion to verify that: i. 24 = 16 ii. 34 = 81
4. a. Expand (1 – 2x)4 b. Now expand (1 + x)(1 – 2x)4
5. Find the first three terms in ascending powers of x in the expansion of (1 + x)4(1 – x)5
6. Using algebraic long division, find the first four terms in the series for 1
1 + x. Now
find the first three terms in the expansion of (1 + x)4
1 + x
7. Write each of the following in the form (a + b)n
a. 1 + 5x + 10x2 + 10x3 + 5x4 + x5
b. 1 – 12x + 60x2 – 160x3 + 240x4 – 192x5 + 64x6
c. 16 + 96x + 216x2 + 216x3 + 81x4
It is sometimes necessary to identify and extract one particular term in the expansion of (a + b)n. In order to do this a theorem must be established for the expansion of (a + b)n which is not tied to Pascal’s Triangle. This requires two basic ‘building blocks’. One is factorials and the other is combinations (both concepts are used in probability).
FactorialsThe product of the first n natural numbers is called n factorial and is written n!
When expanding brackets such as (a + b)5, which is (a + b)(a + b)(a + b)(a + b)(a + b), terms in the expansion are formed by multiplying together one term from each of the five brackets. For example, if b is selected from each bracket, the term b5 results; if b is selected from three of the brackets and a is selected from the other two brackets, then the term a2b3 results. (Since there are two choices of term from each bracket there are 25 = 32 different selections possible and thus 32 terms in the expansion.)
While there is only one way of selecting b5, there are in fact ten different ways of creating the term a2b3, thus in the expansion of (a + b)5 the term in a5 has coefficient 1 and the term in a3b2 has coefficient 10. (Note that the sum of the coefficients in the expansion of (a + b)5 is 1 + 5 + 10 + 10 + 5 + 1 = 32, the total number of different selections possible.)
Fortunately there is a formula that gives us the number of ways of choosing r elements
from a group of n elements, without regard for order. This is written as nr
⎛⎝⎜
⎞⎠⎟
and the number of ways is given by:
nr
⎛⎝⎜
⎞⎠⎟
= n!
(n – r)! r!
The number of ways of choosing three b’s from 5 different brackets is thus:53
⎛⎝⎜
⎞⎠⎟ =
5!(5 – 3)! 3!
[substituting into the formula]
= 5!2! 3!
= 5 × 42 × 1
[cancelling 3!]
= 10 ways.
These selections are often called combinations.
Note: Choosing b from three of the brackets means choosing a from the other two brackets.
Using the formula for the (r + 1)th term, the constant term can be identified without writing out the expansion in full.
Example G
Q. Find the term independent of x in the expansion of ⎛⎝⎜
3x2 – x
⎞ 6
⎠⎟
A. The general (r + 1)th term is
⎛⎝⎜–3x
⎞ r
⎠⎟6r
⎛⎝⎜
⎞⎠⎟(x2)6 – r [using n
r⎛⎝⎜⎞⎠⎟
an – rbr where n = 6, a = x2, b = –3x ]
= (–3)r
xr6r
⎛⎝⎜⎞⎠⎟x
12 – 2r [expanding using index laws]
= 6r
⎛⎝⎜⎞⎠⎟(–3) rx12 – 3r [division law for indices]
For a constant term, the power of x is zero 12 – 3r = 0 r = 4 [solving]
Substituting r = 4 in general term gives
Constant term = 64⎛⎝⎜⎞⎠⎟ (–3)4 x0
= 1 215
Note: An interesting result about finding the sum of the rows of numbers in Pascal’s Triangle was explored in an investigation earlier in this Topic. It is now stated using combinations and the Binomial Theorem. Expanding:
(a + b)6 = 60
⎛⎝⎜
⎞⎠⎟
a6 + 61
⎛⎝⎜
⎞⎠⎟
a5b + 62
⎛⎝⎜
⎞⎠⎟
a4b2 + 63
⎛⎝⎜
⎞⎠⎟
a3b3 + 64
⎛⎝⎜
⎞⎠⎟
a2b4 + 65
⎛⎝⎜
⎞⎠⎟
ab5 + 66
⎛⎝⎜
⎞⎠⎟
b6
Replacing a and b by one, gives:
26 = 60
⎛⎝⎜
⎞⎠⎟
+ 61
⎛⎝⎜
⎞⎠⎟
+ 62
⎛⎝⎜
⎞⎠⎟
+ 63
⎛⎝⎜
⎞⎠⎟
+ 64
⎛⎝⎜
⎞⎠⎟
+ 65
⎛⎝⎜
⎞⎠⎟
+ 66
⎛⎝⎜
⎞⎠⎟
Unit 12.1 Activity 9B: Binomial series1. Find: a. The 4th term in the expansion of (1 + 3x)8
b. The 5th term in the expansion of (1 – 2x)10
c. The 6th term in the expansion of (x + 2y)7
d. The 5th term in the expansion of (x – 2y)8
e. The middle term in the expansion of (2 – x)8
f. The 3rd term in the expansion of (1 – x2)6
2. Find the coefficient of x4 in the expansion of: a. (1 + x)9 b. (1 – 2x)8 c. (1 +
10. Simplify each of the following where possible:
a. x + x2
x + 1 b.
1 + x2
1 + x c.
2n + 1
(2n)n – 1⎛⎝
⎞⎠
÷ 4n + 1
(2n)n – 1⎛⎝
⎞⎠
d. x(x – 5) – 2(3 – 2x)
(x + 1)2 + (x + 1)
11. Solve x + 5 x + 1
+ 23
= –3x
12. Find all values of x for which y = 0 if y = x2 + 2x + 5 – (x + 1)(2x + 2)
(x2 + 2x + 5)2
13. Find the real solutions of the equation x3 – 2x2 – 5x + 6 = 0
14. Complete the following calculations giving exact solutions in the form a + b 3
a. (2 + 3)(4 + 3) b. 14
2 + 315. a. Use the Remainder Theorem, or some other method, to find the remainder
when the cubic function f (x) = x3 – 5x2 + 4x + 6 is divided by (x – 2). b. The cubic equation x3 – 5x2 + 4x + 6 = 0 has one rational and two irrational
solutions. Solve the cubic, leaving your irrational solution in surd form (ie do not give decimal approximations).
16. a. Solve for x: i. (x – 2)2 = 9 ii. x + 8
x = 6 iii. x = x – 2
b. Write 2 log x + log 6x – log 2x in the form log a c. Write (2 + 3)(3 – 2 3) in the form a + b 3
17. Solve the equations for x: a. log
2 x = 3 b. ln (x + 3) = ln x + ln 3 c. ex = 5
18. Solve for x, 2ln x = 3. Give your answer to 3 decimal places.
19. Solve for s the equation s – 1t – 1
– s + 12s
+ 1s = 0
(Hint: there are two solutions, one of which depends on t.)
20. Find all x, y, z satisfying the equations: x = yz, y = xz, z = xy
27. Find the value of the real number p so that the real roots of the equation 2x2 – 12x + p + 2 = 0 are of the form a, a + 2
28. By completing the square, obtain the formula for solving the quadratic equation ax2 + bx + c = 0
29. Determine the value of the term that is independent of x in the expansion
of (x2 + x2 )9
30. a. Use the Binomial Theorem to determine a and b so that (1 + x)n ≈ 1 + ax + bx2 for n ∈ N. b. Use this quadratic approximation to determine the value of (1.03)20. Record