Download Exercise 34: Binomial Theorem

Exercise 34: Binomial Theorem
Objectives: Solve problems, using the Binomial Theorem, for (a + b)N where N belongs
to the set of natural numbers.
Lesson:
Expand and simplify the following binomials:
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
What do you notice about the sum of the exponents in each term?
The sum of the exponents is equal to the power of the binomial expansion.
What do you notice about the exponents on the x-variable? The y-variable?
The exponent of the x-variable begins with the same value as the power of the binomial
and decreases by one in each successive term. The exponent on the y-variable appears in
the second term of the expansion and increases by one until it matches the power of the
binomial.
What do you notice about the number of terms in the expansion?
There are N + 1 terms in the expansion.
What do you notice about the coefficients?
The coefficients are the combinations of the power number beginning with NC0 and
ending at NCN and are symmetrical.
Example 1: Expand (x + y)7 using the binomial theorem.
First, write the pattern for the variables:
(x + y)7 = _x7y0 + _x6y1 +_x5y2 +_x4y3 +_x3y4 +_x2y5 +_x1y6 +_x0y7
Then insert the coefficients:
7C0
7C1
7C2
7C3
7C4
7C5
7C6
7C7
1
7
21
35
35
21
7
1
 (x + y)7 = x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 +7xy6 + y7
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The Binomial Theorem
The binomial theorem states that where n is any positive integer,
n
n
n
n
 n  n 1  n  n
 xy
(x + y)n =   x n    x n 1y    x n  2 y 2    x n  3 y 3  ...  
   y
0
 1
 2
3
 n  1
n
Some additional important observations are:
a) The exponent of the second term of the binomial, in the expansion, is the same as the
number of objects being selected in the combination.
n
That is, we have for the (k+1)st term:   x nk y k
k 
These are always
the same.
b) Or, if we are looking for the kth term, we look to the term:
 n  nk 1 k 1

 x
y
 k  1
c) The sum of the exponents of any term of the expansion if (x + y)n is n.
n
For example:   x nk y k
k 
notice that n-k+k = n
Example 2: Use the binomial theorem to expand each of the following binomials:
(a – b)6 = 6C0a6(-b)0 + 6C1a5(-b)1 + 6C2a4(-b)2+6C3a3(-b)3+6C4a2(-b)4+6C5a1(-b)5 +C6a0(-b)6
= a6  6a5b + 15a4(-b)2 20a3b3 + 15a2b4  6ab5 + b6
(2x – y)4 = 4C0(2x)4(-y)0 + 4C1(2x)3(-y)1 + 4C2(2x)2(-y)2 + 4C3(2x)1(-y)3 + 4C4(2x)0(-y)4
= 116x41 + 48x3(-y) + 64x2y2 + 42x(-y)3 + 1y4
= 16x4  32x3y + 24x2y2  8xy3 + y4
Example 3: Find the simplified form of the eleventh term in the expansion of (x – 2)13.
13 – 10
13C10 x
(-2)10
= 13C10 x3  210
= 286  x3  1024
= 292864x3
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5
2

Example 4: Find the term in the expansion of  x 2   that involves x4.
x

We could solve this by expansion but that is very time consuming. Instead, consider what
we already know:
2
n = 5 a = x2 b =  x
We have:
2
2 5–r
( x )r
5Cr (x )
We can ignore the coefficient for now and simplify:
x10 – 2r  x –r = x 10 – 3r
We would like the exponent on x to be 4:
10 – 3r = 4
6 = 3r
r=2
Therefore:
5C2
2
(x2)5 – 2 ( x )2
4
= 10  x6  x2
= 40 x4
Example 5: Use Pascal’s Triangle to expand (c + d)5
(c + d)5 = 1c5 + 5c4d + 10c3d2 + 10c2d3 + 5cd4 + 1d5
Homework: Exercise 34
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