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7.6: The Binomial Theorem

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A binomial is a polynomial with exactly two terms. The binomial theorem gives a formula for expanding (x+y)n for any positive integer n.

How do we expand a product of polynomials? We pick one term from the first polynomial, multiply by a term chosen from the second polynomial, and then multiply by a term selected from the third polynomial, and so forth. In the special case of (x+y)n, we are selecting either x or y from each of the n binomials x+y to form a product. Some of these products will be identical, hence, we need to collect their coefficients. The expansion of (x+y)3 is demonstrated below.

We find (x+y)3=(x+y)(x+y)(x+y)=xxx+xxy+xyx+xyy+yxx+yxy+yyx+yyy=x3+x2y+x2y+xy2+x2y+xy2+xy2+y3=x3+3x2y+3xy2+y3. What happens when we expand (x+y)n?

If we select y from k copies of the (x+y)s, and x from the other nk copies, their product will be xnkyk. Therefore, in the expansion of (x+y)n, a typical term will be of the form xnkyk, where 0kn. The question is, what is its coefficient in the expansion, after we collect like terms? This coefficient is the number of times the product xnkyk appears when we multiply out (x+y)n in the way described above. It depends on which k copies of the (x+y)s we will choose y from. There are (nk) choices, hence, the product xnkyk appears (nk) times. Thus, the coefficient is (nk). For this reason, we also call (nk) the binomial coefficients.

Theorem 7.6.1 (Binomial Theorem)

For any positive integer n, (x+y)n=nk=0(nk)xnkyk.

Because of the symmetry in the formula, we can interchange x and y. In addition, we also have (nk)=(nnk). Consequently, the binomial theorem can be written in three other forms:

(x+y)n=nk=0(nnk)xnkyk,(x+y)n=nk=0(nk)xkynk,(x+y)n=nk=0(nnk)xkynk.

You need not worry which one to use. They are all the same! This is how to remember these four different forms. In each term, the powers of x and y always add up to n. If the power of one of the two variables is k, where 0kn, then the power of the other must be nk, and we need to multiply the coefficient (nk), which is the same as (nnk), to their product.

When expanding (x+y)n, it may be helpful if you first lay out all the terms xn, xn1y, xn2y2, and so forth. Then you fill in with the binomial coefficients. For instance, to expand (x+y)3, we first list all the terms that we expect fo find:

(x+y)3= _x3+ _x2y+ _xy2+ _y3.

Next we fill in the binomial coefficients:

(x+y)3=(30)x3+(31)x2y+(32)xy2+(33)y3.

Finally, evaluate the binomial coefficients and simplify the result.

(x+y)3=x3+3x2y+3xy2+y3.

In a similar way, we also find (xy)3=x33x2y+3xy2y3. Note the similarity between the two expansions.

Example 7.6.1

Compute (x+y)4.

Solution

Following the steps we outlined above, we find

(x+y)4=(40)x4+(41)x3y+(42)x2y2+(43)xy3+(44)y4=x4+4x3y+6x2y2+4xy3+y4.

Since (n0)=(nn)=1, the expansion always starts with xn and ends with yn.

Example 7.6.2

Compute (xy)4.

Solution

We find

(xy)4=[x+(y)]4=(40)x4+(41)x3(y)+(42)x2(y)2+(43)x(y)3+(44)(y)4=x44x3y+6x2y24xy3+y4.

Take note of the alternating signs in the expansion. This suggests that we could expand (AB)n the exact same way we would with (A+B)n, except that the signs alternate.

We can carry out the expansion by following these steps. First, list all the terms we expect to find

(x+y)4= _x4 _x3y+ _x2y2 _xy3+ _y4.

Next, fill in the signs:

(x+y)4= _x4 _x3y+ _x2y2 _xy3+ _y4,

and then the binomial coefficients:

(x+y)4=(40)x4(41)x3y+(42)x2y2(43)xy3+(44)y4.

Finally, compute the binomial coefficients to finish the expansion.

Example 7.6.3

Expand (2x3y)5.

Solution

The expansion yields (2x)5(51)(2x)4(3y)+(52)(2x)3(3y)2(53)(2x)2(3y)3+(54)(2x)(3y)4(3y)5. Therefore, (2x3y)5=32x5240x4y+720x3y21080x2y3+810xy4243y5.

hands-on Exercise 7.6.1

Use the binomial theorem to expand (3x5y)4.

Example 7.6.4

Find the coefficient of x3 in the expansion of (1+x)102.

Solution

Since (1+x)102=102k=0(102k)xk, the term containing x3 is (1023)x3. Therefore, the coefficient is (1023). Depending on which form of the binomial theorem you use, you may end up with the term (10299)x3. Numerically, this gives us the same coefficient, because (10299)=(10210299)=(1023).
(1023)=171700.

Example 7.6.5

 What is the coefficient of t4 in the expansion of (2+3t)9?

Solution

Since (2+3t)9=9k=0(9k)29k(3t)k, we need k=4. The coefficient is (94)2534=1263281=326,592.

Example 7.6.6

What is the coefficient of t5 in the expansion of (32t)7?

Solution

Since (32t)7=7k=0(7k)37k(2t)k, we need k=5, and the coefficient is (75)32(2)5=(75)3225=21932=6048.

hands-on Exercise 7.6.2

What is the coefficient of t5 in (1+3t)8?

hands-on Exercise 7.6.3

What is the coefficient of t4 in the expansion of (25t)9?

Example 7.6.7

What is the coefficient of t6 in the expansion of (4+5t2)8?

Solution

The general term in the expansion is (8k)48k(5t2)k=(8k)48k5kt2k. Hence, we need k=3, and the coefficient is (83)4553=561024125=7,168,000.

hands-on Exercise 7.6.4

What is the coefficient of t9 in the expansion of (32t3)8?

The constant term in an expansion does not contain any variable. It can be interpreted as the term containing x0.

Example 7.6.8

What is the term with y3 in (3x+5y)8?

Solution

The general term in the expansion is (8k)(3x)8k(5y)k.
Since k=3, the term is (83)(3x)5(5y)3.
This is 56(243)x5(125)y3.

Therefore, the term is 1701000x5y3.

Pascal's Triangle

(See an introduction to Pascal's Triangle in section 7.4)

To compute the binomial coefficients quickly, one may use the Pascal triangle, in which the nth row (n0) consists of the binomial coefficients (nk), where 0kn:

111121133114641151010511615201561

Constructing the Pascal triangle is easy. We generate the rows one at a time. The extreme ends are always 1. Each of the interior entries is the sum of the two entries right above it in the preceding row. For instance, the next row (for n=7) should be

172135352171

Such computations produce the right binomial coefficients, because of the next result.

Theorem 7.6.2 (Pascal's Identitity)

For all integers n and k satisfying 1kn, (nk)=(n1k)+(n1k1).

(Analytic Proof)

It follows from the definition of binomial coefficients that

(n1k1)+(n1k)=(n1)!(k1)!(nk)!+(n1)!k!(nk1)!=(n1)!(k1)!(nk1)!(1nk+1k)=(n1)!(k1)!(nk1)!nk(nk)=n!k!(nk)!.

This completes the proof.

(Combinatorial Proof)

Let A be an n-element set. Then (nk) counts the number of k-element subsets of A. These subsets can be classified according to whether they contain a fixed element, say x. If a subset contains x, then the other k1 elements must be selected from the remaining n1 elements of A. Otherwise, if the subset does not contain x, then all its k elements must be selected from the other n1 elements of A. The numbers of these two kinds of subsets are given by (n1k1) and (n1k), respectively. The theorem now follows immediately by applying the addition principle.

hands-on Exercise 7.6.7

Determine the 8th and the 9th rows in the Pascal’s triangle.

Example 7.6.10

Use the Pascal’s triangle to expand

  1. (CD)5
  2. (2A+5B)3
  3. (3C4B)4
Solution

Draw the values of (nk) from the Pascal triangle directly. The answers are:

  1. (CD)5=C55C4D+10C3D210C2D3+5CD4D5.
  2. (2a+5B)3=8A3+60A2B+150AB2+125B3.
  3. (3C4B)4=81C4432C3B+864C2B2768CB3+256B4.

Summary and Review

  • The binomial theorem can be expressed in four different but equivalent forms.
  • The expansion of (x+y)n starts with xn, then we decrease the exponent in x by one, meanwhile increase the exponent of y by one, and repeat this until we have yn.
  • The next few terms are therefore xn1y, xn2y2, etc., which end with yn.
  • In general, the sum of exponents in x and y is always n. Hence, the general term is xkynk, whose coefficient is (nk).
  • The expansion of (x+y)n and (xy)n look almost identical, except that the signs in (xy)n alternate.

Exercises 

Exercise 7.6.1

Expand (x2y)4

Answer

x48x3y+24x2y232xy3+16y4

Exercise 7.6.2

Find the coefficient of x11y3 in (x+y)14.

Exercise 7.6.3

Find the coefficient of x4y7 in (2xy)11.

Answer

The term is (114)24x4(1)7y7, so the coefficient is 330161=5280.

Exercise 7.6.4

Expand (3ab)5

Exercise 7.6.5

Find the coefficient of  y5 in (2x+3y)7.

Answer

The term is (75)22x235y5, so the coefficient is 214243=20,412.

Exercise 7.6.6

Find the coefficient of y3 in (5x2y)5. 

Exercise 7.6.7

What is the term in (2x+7y)6 with y2?

Answer

The term is (62)24x472y2, so the term is 151649x4y2=11760x4y2

Exercise 7.6.8

What is the term in (x3y)8 with x5?


This page titled 7.6: The Binomial Theorem is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Harris Kwong (OpenSUNY) .

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