Download The Binomial Theorem

The Binomial Theorem
by CHED on June 16, 2017
lesson duration of 10 minutes
under Precalculus
generated on June 16, 2017 at 11:10 pm
Tags: Mathematical Induction
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Generated: Jun 17,2017 07:10 AM
The Binomial Theorem
( 10 mins )
Written By: CHED on July 3, 2016
Subjects: Precalculus
Tags: Mathematical Induction
Resources
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Content Standard
Key concepts of series and mathematical induction and the Binomial Theorem.
Performance Standard
Keenly observe and investigate patterns, and formulate appropriate mathematical statements and prove them using
mathematical induction and/or Binomial Theorem.
Learning Competencies
solve problems using mathematical induction and the Binomial Theorem
determine any term of ???? + ???? ????,
????, where ???? is a positive integer, without expanding
prove the Binomial Theorem
illustrate Pascal’s Triangle in the expansion of ???? + ???? ???? for small positive integral values of ????
Pascal’s Triangle and the Concept of Combination 1 mins
In this lesson, we study two ways to expand (a + b)n, where n is a positive integer. The first, which uses Pascal’s
Triangle, is applicable if n is not too big, and if we want to determine all the terms in the expansion. The second
method gives a general formula for the expansion of (a + b)n for any positive integer n. This formula is useful especially
when n is large because it avoids the process of going through all the coefficients for lower values of n obtained
through Pascal’s Moreover, if only a specific term is required, it can be computed directly using a simple formula.
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Lastly, the theorem can be used to derive and prove some useful and interesting results about sums of combinations.
Teaching Notes Triangle. Calculations with big numbers are required in many of the examples and exercises in this
section. The use of scientific calculators is desirable. You may ask the students to expand these powers using long
multiplication
Consider the following powers of a + b:
We now list down the coefficients of each expansion in a triangular array as follows:
This is part of the Pascal’s Triangle.
Named after the French mathematician Blaise Pascal (1623-1662), some properties of the Pascal’s Triangle are the
following:
(1) Each row begins and ends with 1.
(2) Each row has n + 1 numbers.
(3) The second and second to the last number of each row correspond to the row number.
(4) There is symmetry of the numbers in each row.
(5) The number of entries in a row is one more than the row number (or one more than the number of entries in the
preceding row).
(6) Every middle number after first row is the sum of the two numbers above it. It is the last statement which is useful in
constructing the succeeding rows of the triangle.
It is the last statement which is useful in constructing the succeeding rows of the triangle.
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Example 2.4.1. Use Pascal’s Triangle to expand the expression (2x ? 3y)5.
Solution. We use the coefficients in the fifth row of the Pascal’s Triangle.
Example 2.4.2. Use Pascal’s Triangle to expand (a + b)8.
Solution. We start with the sixth row (or any row of the Pascal’s Triangle that we remember).
Therefore, we get
We observe that, for each n, the expansion of (a + b)n starts with an and the exponent of a in the succeeding terms
decreases by 1, while the exponent of b increases by 1. This observation will be shown to be true in general.
Let us review the concept of combination. Recall that C(n, k) or ( nk ) counts the number of ways of choosing k objects
from a set of n objects. It is also useful to know some properties of C(n, k):
(1) C(n, 0) = C(n, n) = 1,
(2) C(n, 1) = C(n, n – 1) = n, and
(3) C(n, k) = C(n, n – k).
These properties can explain some of the observations we made on the numbers in the Pascal’s Triangle. Recall also
the general formula for the number of combinations of n objects taken k at a time:
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where 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · · n.
You may observe that the value of ( 53 ) and the fourth coefficient in the fifth row of Pascal’s Triangle are the same. In
the same manner, ( 85 ) is equal to the sixth coefficient in the expansion of (a+b)8 (see Example 2.4.2). These observed
equalities are not coincidental, and they are, in fact, the essence embodied in the Binomial Theorem, as you will see in
the succeeding sessions.
Seatwork/Homework 1 mins
1. Use Pascal’s Triangle to expand each expression.
(a) (x – 2y)4
Answer: x4 – 8x3y + 24x2y2 – 32xy3 + 16y4
(b) (2a – b2)3
Answer: 8a3 – 12a2b2 + 6ab4 – b6
(c) (a + b)9
Answer: a9+9a8b+36a7b2 +84a6 b3 +126a5b4+126a4b5+84a3b6+36a2b7+ 9ab8 + b9
2. Compute
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3. Prove:
Answer:
2.4.2. The Binomial Theorem
As the power n gets larger, the more laborious it would be to use Pascal’s Triangle
(and impractical to use long multiplication) to expand (a
(a + b)n. For example,
using Pascal’s Triangle, we need to compute row by row up to the thirtieth row
to know the coefficients of (a
(a + b)30. It is, therefore, delightful to know that it is
possible to compute the terms of a binomial expansion of degree n without going
through the expansion of all the powers less than n.
We now explain how the concept of combination is used in the expansion of
(a + b)n.
When the distributive law is applied, the expansion of (a + b)n consists of
terms of the form ambi, where 0 ? m, i ? n. This term is obtained by choosing
a for m of the factors and b for the rest of the factors. Hence, m + i = n, or
m = n ? i. This means that the number of times the term an?ibi will appear
in the expansion of (a + b)n equals the number of ways of choosing (n
(n ? i) or i
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factors from the n factors, which is exactly C(n, i). Therefore, we have
To explain the reasoning above, consider the case n = 3.
(a + b)3 = (a + b)(a + b)(a + b)
= aaa + aab + aba + abb + baa + bab + bba + bbb
= a3 + 3a2b + 3ab2 + b3
That is, each term in the expansion is obtained by choosing either a or b in each
factor. The term a3 is obtained when a is chosen each time, while a2b is obtained
when a is selected 2 times, or equivalently, b is selected exactly once.
We will give another proof of this result using mathematical induction. But
first, we need to prove a result about combinations.
Pascal's Identity
Proof. The result follows Teaching Notes from the combination formula.
Pascal’s identity explains the method of constructing Pascal’s Triangle, in
which an entry is obtained by adding the two numbers above it. This identity
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is also an essential part of the second proof of the Binomial Theorem, which we
now state.
The Binomial Theorem
For any positive integer n,
Proof. We use mathematical induction.
Part 1
Hence, the formula is true for n = 1.
Part 2. Assume that
We want to show that
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The last expression above follows from Pascal’s Identity.
Therefore, by the Principle of Mathematical Induction,
for any positive integer n.
Terms of a Binomial Expansion 1 mins
2.4.3. Terms of a Binomial Expansion
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We now apply the Binomial Theorem in different examples.
Example 2.4.4. Use the Binomial Theorem to expand (x + y)6.
Solution.
Since the expansion of (a + b)n begins with k = 0 and ends with k = n, the expansion has n + 1 terms. The first term in
the expansion is ( n0 ) an = an, the second term is ( n1 ) anb = nan=1b, the second to the last term is ( nn–1 ) abn–1 =
nabn–1, and the last term is (nn ) bn = bn.
The kth term of the expansion is ( n k?1 ) an–k+1bk–1. If n is even, there is a middle term, which is the (n/2 + 1 ) th term.
If n is odd, there are two middle terms, the (n+1/2) th and (n+1/2 + 1 ) th terms.
The general term is often represented by (nk ) an–kbk. Notice that, in any term, the sum of the exponents of a and b is n.
The combination (nk ) is the coefficient of the term involving bk. This allows us to compute any particular term without
needing to expand (a + b)n and without listing all the other terms.
Example 2.4.5. Find the fifth term in the expansion of ( 2x – ?y )20
Solution. The fifth term in the expansion of a fifth power corresponds to k = 4.
Teaching Notes
To find a specific term in the expansion of (a + b)n, it is important to find the value of k.
Example 2.4.6.
2.4.6. Find the middle term in the expansion of (x/2 + 3y )6 .
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Solution.
Solution. Since there are seven terms in the expansion, the middle term is the fourth term (k = 3), which is
Example 2.4.7. Find the term involving x (with exponent 1) in the expansion of ( x2 – 2y/x )8 .
Solution. The general term in the expansion is
The term involves x if the exponent of x is 1, which means 16 ? 3k = 1, or k = 5. Hence, the term is
Seatwork/Homework 1 mins
1. Use the Binomial Theorem to expand (2a ? b2)5.
Answer:
2. Find the two middle terms in the expansion of ( x1/3+ 2/y )11 .
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Answer: There are 12 terms in the expansion, so the two middle terms are the 6th (corresponding to k = 5) and the 7th
(corresponding to k = 6) terms.
3. Find the constant term in the expansion of ( x3 2 + 3 x2 )10 .
Answer: The general term is
The constant term contains x0, which means 30 – 5k = 0, or k = 6.
Approximation and Combination Identities 1 mins
We continue applying the Binomial Theorem.
*Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in the expansion of (1 – 0.2)8. Compare your
answer with the calculator value.
(2) Use 5 terms in the binomial expansion to approximate (0.8)8. Is there an improvement in the approximation?
Solution.
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The calculator value is 0.16777216, so the error is 0.35222784.
The error is 0.01622784, which is an improvement on the previous estimate.
Example 2.4.9. Use the Binomial Theorem to prove that, for any positive integer n,
Solution.
Solution. Set a = b = 1 in the expansion of (a + b)n. Then
Example 2.4.10. Use the Binomial Theorem to prove that
Solution. Let a = 1 and b = –1 in the expansion of (a + b)100. Then
If k is even, then (–1)k = 1. If k is odd, then (–1)k = –1. Hence, we have
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Therefore, after transposing the negative terms to other side of the equation, we obtain
Approximation and Combination Identities 1 mins
We continue applying the Binomial Theorem.
*Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in the expansion of (1 – 0.2)8. Compare your
answer with the calculator value.
(2) Use 5 terms in the binomial expansion to approximate (0.8)8. Is there an improvement in the approximation?
Solution.
The calculator value is 0.16777216, so the error is 0.35222784.
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The error is 0.01622784, which is an improvement on the previous estimate.
Example 2.4.9. Use the Binomial Theorem to prove that, for any positive integer n,
Solution.
Solution. Set a = b = 1 in the expansion of (a + b)n. Then
Example 2.4.10. Use the Binomial Theorem to prove that
Solution. Let a = 1 and b = –1 in the expansion of (a + b)100. Then
If k is even, then (–1)k = 1. If k is odd, then (–1)k = –1. Hence, we have
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Therefore, after transposing the negative terms to other side of the equation, we obtain
Seatwork/Homework 1 mins
*1. Approximate (1.9)10using the first three terms in the expansion of (2 – 0.1)10, and find its error compared to the
calculator value.
Answer:
Calculator value = 613.1066258
Error from the calculator value = 14.09337422
2. Prove that, for any positive integer n,
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Seatwork/Homework 1 mins
*1. Approximate (1.9)10using the first three terms in the expansion of (2 – 0.1)10, and find its error compared to the
calculator value.
Answer:
Calculator value = 613.1066258
Error from the calculator value = 14.09337422
2. Prove that, for any positive integer n,
Exercises 1 mins
1. Use the Binomial Theorem to expand each expression.
(a) (x – 2)5
Answer: x5 – 10x4 + 40x3 – 80x2 + 80x –32
(b) ( x + 1/y )7
Answer: x7 + 7x6/y + 21x5/y2 + 35x4/y3 + 35x3/y4 + 21x2/y5 + 7x/y6 + 1/y7
(c) ( 3 – 1/2 )4
Answer: 81 – 3(27) ( 1/2 ) + 3(9) ( 1/4 ) – 1/8 = 377/8
2. Without expanding completely, compute the indicated term(s) in the expansion of the given expression.
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*3. Approximate (1.1)10 by using the first 4 terms in the expansion of (1 + 0.1)10. Compare your answer with the
calculator result.
Answer: 2.57, with an error of 0.0237424601 from the calculator value of 2.59374246
4. Use the Binomial Theorem to prove that
Hint: Expand (1 + 2)n.
5. Use the Binomial Theorem to prove that
Hint: Expand (1 – 250.
Exercises 1 mins
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1. Use the Binomial Theorem to expand each expression.
(a) (x – 2)5
Answer: x5 – 10x4 + 40x3 – 80x2 + 80x –32
(b) ( x + 1/y )7
Answer: x7 + 7x6/y + 21x5/y2 + 35x4/y3 + 35x3/y4 + 21x2/y5 + 7x/y6 + 1/y7
(c) ( 3 – 1/2 )4
Answer: 81 – 3(27) ( 1/2 ) + 3(9) ( 1/4 ) – 1/8 = 377/8
2. Without expanding completely, compute the indicated term(s) in the expansion of the given expression.
*3. Approximate (1.1)10 by using the first 4 terms in the expansion of (1 + 0.1)10. Compare your answer with the
calculator result.
Answer: 2.57, with an error of 0.0237424601 from the calculator value of 2.59374246
4. Use the Binomial Theorem to prove that
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Hint: Expand (1 + 2)n.
5. Use the Binomial Theorem to prove that
Hint: Expand (1 – 2)50.
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