MTH5P1: Binomial Theorem

This unit explains Binomial Theorem

Suppose you need to calculate the amount of interest you will get after 5 years on a sum of money that you have invested at the rate of 15% compound interest per year. Or suppose we need to find the size of the population of a country after 10 years if we know the annual growth rate. A result that will help in finding these quantities is the binomial theorem.

This theorem, as you will see, helps us to calculate the rational powers of any real binomial expression, that is, any expression involving two terms.

The binomial theorem, was known to Indian and Greek mathematicians in the 3rd century B.C. for some cases. The credit for the result for natural exponents goes to the Arab poet and mathematician Omar Khayyam (A.D. 1048-1122). Further generalisation to rational exponents was done by the British mathematician Newton (A.D. 1642-1727).

There was a reason for looking for further generalisation, apart from mathematical interest. The reason was its many applications. Apart from the ones we mentioned at the beginning, the binomial theorem has several applications in probability theory, calculus, and in approximating numbers like (1.02)7, 31/5, etc. We shall discuss a few of them in this lesson.

Before discussing Binomial Theorem, we shall introduce the concept of Principle of Mathematical Induction, which we shall be using in proving the Binomial Theorem for a positive integral index. This principle is also useful in making generalisations from particular statements/results.

WHAT IS A STATEMENT ?

In your daily interactions, you must have made several assertions in the form of sentences. Of
these assertions, the ones that are either true or false are called statement or propositions.
For instance,
“I am 20 years old” and “If x = 3, then are statements, but ‘When will you leave ?’ And ‘How wonderful!’ are not statements.
Notice that a statement has to be a definite assertion which can be true or false, but not both.
For example,  ‘x – 5 = 7’ is not a statement, because we don’t know what x, is. If x = 12, it is true, but if x = 5, ‘it is not true. Therefore, ‘x – 5 = 7’ is not accepted by mathematicians as a statement.
But both ‘x – 5 = 7Þx = 12’ and x – 5 = 7 for any real number x’ are statements, the first one true and the second one false.

Example 8.1 Which of the following sentences is a statement ?
(i) India has never had a woman President.
(ii) 5 is an even number.

Solution : (i) and (ii) are statements, (i) being true and (ii) being false. (iii) is not a statement, since we can not determine whether it is true or false, unless we know the range of values that x and y can take.

Now look at (iv). At first glance , you may say that it is not a statement, for the very same reasons that (iii) is not. But look at (iv) carefully. It is true for any value of a and b. It is an identity. Therefore, in this case, even though we have not specified the range of values fora and b, (iv) is a statement.

Some statements, like the one given below are about natural numbers in general. Let us look at the statement given below :

This involves a general natural number n. Let us call this statement P (n) [P stands for proposition].

Similarly, P (2) would be the statement

Let us look at some examples to help you get used to this notation.

Example 8.2 If P (n) denotes 

Solution : Replacing n by 1, k and k + 1, respectively in P (n), we get

Example 8.3 If P (n) is the statement

Solution : To write P(1), the terms on the left hand side (LHS) of P(n) continue till 3x 1 – 2, i.e., 1. So, P (1) will have only one term in its LHS, i.e., the first term.

Also, the right hand side (RHS)ofP(1) =

THE PRINCIPLE OF MATHEMATICAL INDUCTION

In your daily life, you must be using various kinds of reasoning depending on the situation you are faced with. For instance, if you are told that your friend just had a child, you would know that it is either a girl or a boy. In this case, you would be applying general principles to a particular case. This form of reasoning is an example of deductive logic.

Now let us consider another situation. When you look around, you find students who study regularly, do well in examinations, you may formulate the general rule (rightly or wrongly) that “any one who studies regularly will do well in examinations”. In this case, you would be formulating a general principle (or rule) based on several particular instances. Such reasoning is inductive, a process of reasoning by which general rules are discovered by the observation and consideration of several individual cases. Such reasoning is used in all the sciences, as well as in
Mathematics.

Mathematical induction is a more precise form of this process. This precision is required because a statement is accepted to be true mathematically only if it can be shown to be true for each and every case that it refers to. The following principle allows us to check if this happens.

The Principle of Mathematical Induction:
Let P(n) be a statement involving a natural number n. If
(i) it is true for n = 1, i.e., P(1) is true; and

(ii) assuming and P(k) to be true, it can be proved that P(k+ 1) is true; then P(n) must be true for every natural number n.

Note that condition (ii) above does not say that P(k) is true. It says that whenever P(k) is true, then P( k + 1) is true’.
Let us see, for example, how the principle of mathematical induction allows us to conclude that P(n) is true for n = 11.

By (i) P(1) is true. As P(1) is true, we can put k = 1 in (ii), So P(1 + 1), i.e., P(2) is true. As P(2) is true, we can put k = 2 in (ii) and conclude that P(2 + 1), i.e., P(3) is true. Now put k = 3 in (ii), so we get that P(4) is true. It is now clear that if we continue like this, we shall get that P(11) is true.

It is also clear that in the above argument, 11 does not play any special role. We can prove that P(137) is true in the same way. Indeed, it is clear that P(n) is true for all n > 1.

Let us now see, through examples, how we can apply the principle of mathematical induction to prove various types of mathematical statements.

Example 8.4 Prove that

where n is a natural number.

Solution:We have

Therefore, which is true,.

Therefore, P(1) is true.
Let us now see, if P(k + 1) is true whenever P(k) is true.
Let us, therefore, assume that P(k) is true, i.e.,

It will be true, if we can show that LHS = RHS

So, P(k + 1) is true, if we assume that P(k) is true.
SinceP(1) is also true, both the conditions of the principle of mathematical induction are fulfilled, we conclude that the given statement is true for every natural number n.

As you can see, we have proved the result in three steps – the basic step [i.e., checking (i)], the Induction step [i.e., checking (ii)], and hence arriving at the end result.

Example 8.5 Prove that

where n is a natural number.

Therefore, P(k + 1) is true.
Hence, by the principle of mathematical induction, the given statement is true for every natural number n.

Example 8.6 For ever natural number n, prove that is divisible by (x + y),

Solution:Let us see if we can apply the principle of induction here. Let us call P(n) the statement

Then P(1) is ‘ is divisible by (x + y)’, i.e., ‘(x + y) is divisible by (x + y)’, which is true.
Therefore, P(1) is true.

Let us now assume that P(k) is true for some natural number k, i.e., is divisible by (x + y).

This means that for some natural number t,

We wish to prove that P (k +1) is true, i.e., is divisible by (x+y)’ is true.
Now.

which is divisible by of (x + y).
Thus, P (k+1) is true.
Hence, by the principle of mathematical induction, the given statement is true for every natural number n.

Example 8.7 Prove that n for every natural number n.

Hence, by the principle of mathematical induction, the given statement is true for every natural number n.

Sometimes, we need to prove a statement for all natural numbers greater than a particular natural number, say a (as in Example 8.8 below). In such a situation, we replace P(1) by P(a +1) in the statement of the principle.

Example 8.8 Prove that

where n is a natural number.

Solution: For let us call the following statement

Since we have to prove the given statement for the first relevant statement isP(3). We, therefore, see whether P(3) is true.

So, P (3) is true.

We wish to prove that P (k + 1) is true.

Therefore, P(k + 1) is true.
Hence, by the principle of mathematical induction, the given statement is true for every natural number 

Example 8.9 Using principle of mathematical induction, prove that

is a natural number for all natural numbers n.

Solution :

THE BINOMIAL THEOREM FOR A NATURAL EXPONENT

You must have multiplied a binomial by itself, or by another binomial. Let us use this knowledge to do some expansions. Consider the binomial (x + y). Now,

In each of the equations above, the right hand side is called the binomial expansion of the left hand side.
Note that in each of the above expansions, we have written the power of a binomial in the expanded form in such a way that the terms are in descending powers of the first term of the binomial (which is x in the above examples). If you look closely at these expansions, you would also observe the following:
1. The number of terms in the expansion is one more than the exponent of the binomial. For example, in the expansion of the number of terms is 5.
2. The exponent of x in the first term is the same as the exponent of the binomial, and the exponent decreases by 1 in each successive term of the expansion.

3. The exponent of y in the first term is zero  The exponent of y in the second term is 1, and it increases by 1 in each successive term till it becomes the exponent of the binomial in the last term of the expansion.
4. The sum of the exponents of x and y in each term is equal to the exponent of the binomial. For example, in the expansion of  the sum of the exponents of x and y in each term is 5.
If we use the combinatorial coefficients, we can write the expansion as

More generally, we can write the binomial expansion of  where n is a positive integer, as given in the following theorem. This statement is called the binomial theorem for a natural (or positive integral) exponent.

Theorem 8.1

where nÎN and x, y Î R.
Proof :Let us try to prove this theorem, using the principle of mathematical induction.
Let statement (A) be denoted byP(n), i.e.,

Let us examine whether P(1) is true or not.
From (i), we have

Assuming that P(k) is true, if we prove that P(k + 1) is true, then P(n) holds, for all n. Now,

which shows that P(k +1) is true.
Thus, we have shown that (a) P(1) is true, and (b) if P(k) is true, then P(k+1) is also true.
Therefore, by the principle of mathematical induction,P(n) holds for any value of n. So, we have proved the binomial theorem for any natural exponent.
This result is supported to have been proved first by the famous Arab poet Omar Khayyam, though no one has been able to trace his proof so far.
We will now take some examples to illustrate the theorem.

Example 8.10 Write the binomial expansion of 

Solution :Here the first term in the binomial isx and the second term is 3y. Using the binomial theorem, we have

Example 8.11 Expand in terms of powers of a, where a is a real number.
Solution : Taking x = 1 and y = a in the statement of the binomial theorem, we have

(B) is another form of the statement of the binomial theorem.
The theorem can also be used in obtaining the expansions of expressions of the type

 

 

 

Let us illustrate it through an example.
Example 8.12 Write the expansion of 

Solution :We have :

Example 8.13 The population of a city grows at the annual rate of 3%. What percentage increase is expected in 5 years ? Give the answer up to 2 decimal places.

Solution : Suppose the population is a at present. After 1 year it will be

Using the binomial theorem, and ignoring terms involving more than 3 decimal places, we get

Example 8.14 Using binomial theorem, evaluate

GENERAL AND MIDDLE TERMS IN A BINOMIAL EXPANSION

Let us examine various terms in the expansion (A) of i.e., in

We observe that

generally referred to as the general term of the binomial expansion.
Let us now consider some examples and find the general terms of some expansions.

Example 8.15 Find the term in the expansion of where n is a natural number. Verify your answer for the first term of the expansion.
Solution :The general term of the expansion is given by :

This verifies that the expression for  correct for r + 1 =1.

Example 8.16 Find the fifth term in the expansion of

Solution : Using here which gives r +1 = 5, i.e., r = 4.

Now that you are familiar with the general term of an expansion, let us see how we can obtain the middle term(or terms) of a binomial expansion. Recall that the number of terms in a binomial expansion is always one more than the exponent of the binomial. This implies that if the exponent is even, the number of terms is odd, and if the exponent is odd, the number of terms is even. Thus, while finding the middle term in a binomial expansion, we come across two cases:
Case 1 : When n is even.
To study such a situation, let us look at a particular value ofn, say n =6. Then the number of terms in the expansion will be 7. From Fig. 8.1, you can see that there are three terms on either side of the fourth term.

In general, when the exponent n of the binomial is even, there are  terms on either side of the term. Therefore, the term is the middle term.

Case 2:When n is odd
Let us take n =7 as an example to see what happens in this case. The number of terms in the expansion will be 8. Looking at Fig. 8.2, do you find any one middle term in it? There is not. But we can partition the terms into two equal parts by a line as shown in the figure. We call the terms on either side of the partitioning line taken together, the middle terms. This is because there are an equal number of terms on either side of the two, taken together.

Thus, in this case, there are two middle terms, namely, the fourth,

Similarly, if n =13, then the  and the  terms, i.e., the 7th and 8th terms are two middle terms, as is evident from Fig. 8.3.
From the above, we conclude that

When the exponent n of a binomial is an odd natural number, then the and terms are two middle terms in the corresponding binomial expansion.
Let us now consider some examples.

Example 8.17 Find the middle term in the expansion of 

Solution :Here n = 8 (an even number).

Therefore, the  the 5th term is the middle term.

Example 8.18 Find the middle term(s) in the expansion of 

Solution :Here n =9 (an odd number). Therefore, the  are middle terms. i.e. terms. i.e.

are middle terms.

For finding  putting r =4 and r = 5 in the general term

BINOMIAL THEOREM FOR RATIONAL EXPONENTS

So far you have applied the binomial theorem only when the binomial has been raised to a power which is a natural number. What happens if the exponent is a negative integer, or if it is a fraction? We will state the result that allows us to still have a binomial expansion, but it will have infinite terms in this case.
The result is a generalised version of the earlier binomial theorem which you have studied.
Theorem 8.2. The Binomial Theorem for a Rational Exponent.

If r is a rational number, and x is a real number such that |x| < 1, then

We will not prove this result here, as it is beyond the scope of this course. In fact, even Sir Issac Newton, who is credited with stating this generalisation, stated it without proof in two letters, written in A.D. 1676. The proof was developed later, by other mathematicians, in stages. Among those who contributed to the proof of this theorem were English mathematician
Colin Maclaurin (A.D. 1698-1746) for rational values of r, Giovanni Francesco, M.M. Salvemini (A.D. 1708-1783) and the German mathematician Abraham G. Kasther (A.D. 1719 – 1800) for integral values of r, the Swiss mathematician Leonhard Euler (AD 1707-1783) for fractional exponents and the Norwegian mathematician Neils Henrik Abel (1802-1829) for complex exponents. Let us consider some examples to illustrate the theorem.

Example 8.19 Write the expansion of

Solution :Here r = –1 [with reference to (D) above].
Therefore,

Similarly, you can write the expansion 

Note the above expansions. In case of  all the terms have positive and negative signs alternate, while in the case of  all the terms have positive sign.
You may have also observed the following points about the binomial expansion (D) in general;

1. If r is a natural number, then (C) and (D) coincide for the case 

2. Note that etc. Thus, the coefficients 1,  in (D) look like combinatorial coefficients.,

However, recall that  for natural numbers r and whole numbers only.
Therefore,

have no meaning in the present context.

3. The expression (D) will have an infinite number of terms.
4. The sum of the series on the RHS of (D) may not be meaningful if x >1.
For example, if we put x = 2 in Example 1, we have

which is clearly false.

Therefore, for (D) to hold, it is necessary that |x| < 1.
Let us look at some more examples of this binomial expansion.

Example 8.20 Expand  where r is a rational number and

The result we have just obtained in Example 8.20 is another form of the binomial theorem for a rational exponent.Let us restate it formally.

Consequently, we have the following result:

For a rational number r, an expression like  can be expanded in two different ways, depending on whether

Note that in (i), we have obtained the expansion in ascending powers of y while in (ii), we have obtained the expansion in ascending powers of x.

USE OF BINOMIAL THEOREM IN APPROXIMATIONS

As you have seen, the binomial expansions sometime have infinitely many terms. In such cases, for further calculations; an approximate value involving only the first few terms may be enough for us. Let us illustrate some situations in which we find the approximate values.
Example 8.22 Find the cube root of 1.03 up to three decimal places.

Solution : We want to find  up to three decimal places.

Now, we need to approximate the value up to three decimal places. Since a non-zero digit in the fourth decimal place may affect the digit in the third place in the process of rounding off, we need to consider those terms in the expansion which produce a non-zero digit in the first, second, third or fourth decimal place.
Therefore, we can take the sum of the first three terms in the Expansion (i), and ignore the rest.

Now, the digit after the third decimal place is greater than 5, so we have increased the third decimal place by 1.
Thus, the cube root of 1.03, up to three decimal places, is 1.010.
Example 8.23 Assuming y to be so small that y2 and higher powers of y can be neglected, find the value of 

Solution : Note that y is very small. So, we can assume that  Then, using the binomial theorem, we get

Thus, the given product is approximately

 

ASSIGNMENT : Binomial Theorem Assignment MARKS : 30  DURATION : 60 minutes

 

Welcome to FAWE

STEM Elearning

We at FAWE have built this platform to aid learners, trainers and mentors get practical help with content, an interactive platform and tools to power their teaching and learning of STEM subjects, more

How to find your voice as a woman in Africa

top
© FAWE, Powered by: Yaaka DN.
X