THE BINOMIAL THEOREM4.1
4.1.2 THE GENERAL TERM
We have already seen the relationship between Pascal’s triangle and its combinatorial equivalent.
From this relationship we were able to produce the general expansion for . That is,
Where the first term, =
It is common in examinations for questions to only ask for a part of an expansion. This is because the previous examples are time consuming to complete.
The fifth term is given by . Using the general term, this means that . For this expansion we have that n = 10, therefore,
Therefore the fifth term is .
x+a
The (r +1)th term is also know as the general term. That is tr+1 n
r
xn–rar
=
Find the 5th term in the expansion , when expanded in descending powers of x.
101 The general term for this expansion is given by
=
Seeing as we want the term involving we equate the power of x in the general term to 12:
So, .
Then, . So, the coefficient of is in fact 112.
In this case we want the term independent of x, that is, the term that involves . Again, we first find an expression for the general term,
Notice how we had to separate the constants and the x term.
Next, we equate the power of x in the expansion to 0: .
We therefore want .
So, the term independent of x is 240.
i. can be approximated using the expansion of and an x value of 0.01 using the first three terms of the series.
ii. The correct answer can be found using the power key of a calculator and is Find the coefficient of x12 in the expansion (x2–2)8.
Write the expansion of
i. Use the first three terms of the series to approximate
ii. Find the absolute, relative and percentage errors in making this approximation.
1+x
102
1.061520150601. It is probable that most calculators will not be able to display the 12 decimal places of the full answer. An answer such as 1.0615201506 is sufficient for the remaining part of this problem
The absolute error =
The relative error
The percentage error %
1. Find the terms indicated in the expansions of the following expressions:
Expression Term
2. Find the coefficients of the terms indicated in the expansions of the following expressions:
Expression Term
3. Use the first three terms in the expansion of to find an approximate value for . Find the percentage error in using this approximation.
4. (i) Write the expansion of .
(ii) Use the first three terms of the expansion to approximate . 1.0615–1.0615201 ≈2.015×10–5
103
(iii) Find the absolute error in this approximation.
(iv) Find the percentage error in this approximation.
5. Find the coefficient of in the expansion of .
6. Find the constant term in the expansion of .
7. Find the constant term in the expansion of .
8. Find the term independent of x in the expansion of .
9. Find the term independent of x in the expansion of .
10. In the expansion of , where a is a non-zero constant, the coefficient of the term in is ‘–9’ times the coefficient in . Find the value of the constant a.
11. If the coefficient of the in the expansion of is 90, find n.
12. Three consecutive coefficients in the expansion of are in the ratio 6 : 14 : 21.
Find the value of n.
13. Find the independent term in the following expansions
(a) (b)
14. In the expansion of the first term is 1, the second term is and the third term is . Find the values of a and n.
15. In the expansion of the coefficient of is –9 and there is no term.
Find a and b.
16. By considering the expansion of , prove that:
(a)
104
Note: This proof by induction is outside the scope of the syllabus.
A formal statement of the binomial expansion is:
=
The binomial theorem for positive integral index may be proved using mathematical induction.
A preliminary result from combinatorics is required, namely . We leave its proof as an exercise. We can now move on to the main induction proof:
(i) Check the case k = 1: , which is true.
(ii) Assume the theorem is true for n = k:
(iii) Look at the n = k + 1. This involves multiplying each term in the expansion from (ii) first by a and then by b. To see what happens to the general term, it is a good idea to look at two consecutive terms in the middle of the expansion from (ii):
When this expansion has been multiplied by a, the result is:
and when it is multiplied by b, the result is:
The expansion of begins with , which is correct. The
expansion ends with , which is also correct.
It now remains to prove that the general term in the middle of the expansion is also correct.
Lining up like terms from the two parts of the expansion gives:
The general term is:
[Using the combinatorial result given at the start of this section].
(iv) We can conclude that the binomial theroem gives the correct expansion for n = 1 from part (i). Part (iii) indicates that the theorem gives the correct expansion for an index of 2, 3 etc.
Hence the theroem holds for all positive integral indices.
PROOF 4.2
a+b
( )k = nC0akb0+nC1ak–1b1+…+nCrak–rbr+…+nCka0bk nCran–rbr
r=0
∑
nnC
r+nCr–1 = n+1Cr
a+b
( )1 = 1C0a1+1C0b1 = a+b
a+b
( )k = kC0akb0+kC1akb1+…+kCrak–rbr+…+kCka0bk
a+b
( )k = kC0akb0+…+kCr–1ak–r+1br–1+kCrak–rbr+…+kCka0bk
kC
0ak+1b0+…+kCr–1ak–r+2br–1+kCrak–r+1br+…+kCka1bk
kC
0akb1+…+kCr–1ak–r+1br+kCrak–rbr+1+…+kCka0bk+1 a+b
( )k+1 kC0ak+1b0 = ak+1b0 = ak+1
kC
ka0bk+1 = a0bk+1 = bk+1
kC
r–1ak–r+2br–1+kCrak–r+1br
kC
r–1ak–r+1br–1+kCrak–rbr+1
kC
rak–r+1br+kCr–1ak–r+1br = [kCr+kCr–1](ak–r+1br)
k+1C
rak–r+1br
=
Functions and Relations – CHAPTER
5
5.1.1 RELATIONS
Consider the relationship between the weight of five students and their ages as shown below.
We can represent this information as a set of ordered pairs. An age of 10 years would correspond to a weight of 31kg. An age of 16 years would correspond to a weight of 53kg and so on.
This type of information represents a relation between two sets of data. This information could then be represented as a set of ordered pairs,
The set of all first elements of the ordered pair is called the domain of the relation and is referred to as the independent variable. The set of all second elements is called the range and is referred to as the dependent variable.
For the above example, the domain =
and the range = .
Notice that and are not the same! This is because the ordered pair provides the correct relation between age and weight, i.e., at age 10 years the weight of the student is 31kg. On the other hand, the ordered pair would be informing us that at age 31 years the weight of the student is 10kg!
(a) The domain is the set of all first elements, i.e., {0, 1, 2, 3, 4, 5}.
Age (years) Weight (kg)
10 31
Set of ordered pairs {(x, y)}
Domain range defining
Determine the domain and range for each of the following relations:
(a) .
MATHEMATICS Standard Level
The letter ‘X’ is often used to denote the domain and the letter ‘Y’ to denote the range. For part (a) this means that we could write X = {0, 1, 2, 3, 4, 5} and Y = {0, 1, 4, 9, 16, 25} and for (b) we could write X = {–3, –1, 2, –2} and Y = {4, 0, –2, 2}.
This is a convention, nothing more.
Rather than giving a verbal description of how the independent variable and the dependent variable are related, it is much clearer to provide a mathematical rule that shows how the elements in the range relate to the elements in the domain.
(a) The domain of this relation is given by the x–values, i.e., {0, 1, 2, 3, 4}. We can therefore substitute these values into the equation y = x + 2 and determine their corresponding y–values. This will provide the range of the relation.
Substituting we have,
, and so on.
This produces a set of y–values {2, 3, 4, 5, 6} that defines the range.
(b) The set of ordered pairs would be {(0, 2), (1, 3), (2, 4), (3, 5), (4, 6)}.
Notice that we can describe the set of ordered pairs more formally as:
. which is read as;
“The set of ordered pairs x and y, such that y = x + 2, where x is an element of the set of values {0, 1, 2, 3, 4}.”
The information in Example 5.2 can be displayed in different ways. Both those shown below are visual displays – they show the mappings in different ways:
A relation is defined by the rule .
(a) Determine the range of this relation.
(b) Express this relation as a set of ordered pairs.
y = x+2 where x, ∈{0 1 2 3 4, , , , }
E
XAMPLE5.2S
o l u t i o nx = 0⇒y = 0+2 =2 x = 1⇒y = 1+2 =3 x = 2⇒y = 2+2 =4
x y
( , ):y=x+2 x, ∈{0 1 2 3 4, , , , }
{ }
Mapping diagram
The mapping diagram below displays which y–value corresponds to a given x–value.
However it is often not easy to see the ‘pattern’
Cartesian Plane
The Cartesian plane is made up of a horizontal axis (independent variable, X) and a vertical axis (dependent variable,Y).
We plot the points on the grid, so that (3, 5) is 3 0 2
1 3 3 5 4 6 2 4
Domain (X) Range (Y)
1 23 45 6
1 2 3 4
0 X
Y
MM Chapter 005 [105 – 166] Page 106 Sunday, April 1, 2007 10:53 PM
Functions and Relations – CHAPTER
5
Notice that in the mapping diagram that uses the Cartesian plane, we have not joined the points together in a straight line. This is because the domain specifies that the only values of x that can be used must be from the set {0, 1, 2, 3, 4}, and so a value such as x = 2.4 cannot be used.
Both these visual representations are useful in displaying which values in the domain generate a given value in the range. However, the Cartesian plane more readily gives a quick overview of what the underlying relationship between the two variables is. It is very easy (and quick) to see that as the x–values increase, so too do the y–values. We can do this by simply looking at the points on the graph and observing the ‘trend’ without really concerning ourselves with what the actual values are.
We now provide a formal definition of the Cartesian plane and a relation.