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 n−k copies, their product will be xn−kyk. Therefore, in the expansion of (x+y)n, a typical term will be of the form xn−kyk, where 0≤k≤n. The question is, what is its coefficient in the expansion, after we collect like terms? This coefficient is the number of times the product xn−kyk 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 xn−kyk 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=n∑k=0(nk)xn−kyk.
Because of the symmetry in the formula, we can interchange x and y. In addition, we also have (nk)=(nn−k). Consequently, the binomial theorem can be written in three other forms:
(x+y)n=n∑k=0(nn−k)xn−kyk,(x+y)n=n∑k=0(nk)xkyn−k,(x+y)n=n∑k=0(nn−k)xkyn−k.
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 0≤k≤n, then the power of the other must be n−k, and we need to multiply the coefficient (nk), which is the same as (nn−k), to their product.
When expanding (x+y)n, it may be helpful if you first lay out all the terms xn, xn−1y, xn−2y2, 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 (x−y)3=x3−3x2y+3xy2−y3. 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 (x−y)4.
- Solution
-
We find
(x−y)4=[x+(−y)]4=(40)x4+(41)x3(−y)+(42)x2(−y)2+(43)x(−y)3+(44)(−y)4=x4−4x3y+6x2y2−4xy3+y4.
Take note of the alternating signs in the expansion. This suggests that we could expand (A−B)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 (2x−3y)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, (2x−3y)5=32x5−240x4y+720x3y2−1080x2y3+810xy4−243y5.
hands-on Exercise 7.6.1
Use the binomial theorem to expand (3x−5y)4.
Example 7.6.4
Find the coefficient of x3 in the expansion of (1+x)102.
- Solution
-
Since (1+x)102=102∑k=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)=(102102−99)=(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=9∑k=0(9k)29−k(3t)k, we need k=4. The coefficient is (94)25⋅34=126⋅32⋅81=326,592.
Example 7.6.6
What is the coefficient of t5 in the expansion of (3−2t)7?
- Solution
-
Since (3−2t)7=∑7k=0(7k)37−k(−2t)k, we need k=5, and the coefficient is (75)32⋅(−2)5=−(75)32⋅25=−21⋅9⋅32=−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 (2−5t)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)48−k(5t2)k=(8k)48−k⋅5kt2k. Hence, we need k=3, and the coefficient is (83)45⋅53=56⋅1024⋅125=7,168,000.
hands-on Exercise 7.6.4
What is the coefficient of t9 in the expansion of (3−2t3)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)8−k(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 (n≥0) consists of the binomial coefficients (nk), where 0≤k≤n:
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 1≤k≤n, (nk)=(n−1k)+(n−1k−1).
- (Analytic Proof)
-
It follows from the definition of binomial coefficients that
(n−1k−1)+(n−1k)=(n−1)!(k−1)!(n−k)!+(n−1)!k!(n−k−1)!=(n−1)!(k−1)!(n−k−1)!(1n−k+1k)=(n−1)!(k−1)!(n−k−1)!⋅nk(n−k)=n!k!(n−k)!.
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 k−1 elements must be selected from the remaining n−1 elements of A. Otherwise, if the subset does not contain x, then all its k elements must be selected from the other n−1 elements of A. The numbers of these two kinds of subsets are given by (n−1k−1) and (n−1k), 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
- (C−D)5
- (2A+5B)3
- (3C−4B)4
- Solution
-
Draw the values of (nk) from the Pascal triangle directly. The answers are:
- (C−D)5=C5−5C4D+10C3D2−10C2D3+5CD4−D5.
- (2a+5B)3=8A3+60A2B+150AB2+125B3.
- (3C−4B)4=81C4−432C3B+864C2B2−768CB3+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 xn−1y, xn−2y2, etc., which end with yn.
- In general, the sum of exponents in x and y is always n. Hence, the general term is xkyn−k, whose coefficient is (nk).
- The expansion of (x+y)n and (x−y)n look almost identical, except that the signs in (x−y)n alternate.
Exercises
Exercise 7.6.1
Expand (x−2y)4
- Answer
-
x4−8x3y+24x2y2−32xy3+16y4
Exercise 7.6.2
Find the coefficient of x11y3 in (x+y)14.
Exercise 7.6.3
Find the coefficient of x4y7 in (2x−y)11.
- Answer
-
The term is (114)24⋅x4(−1)7y7, so the coefficient is 330⋅16⋅−1=−5280.
Exercise 7.6.4
Expand (3a−b)5
Exercise 7.6.5
Find the coefficient of y5 in (2x+3y)7.
- Answer
-
The term is (75)22⋅x2⋅35y5, so the coefficient is 21⋅4⋅243=20,412.
Exercise 7.6.6
Find the coefficient of y3 in (5x−2y)5.
Exercise 7.6.7
What is the term in (2x+7y)6 with y2?
- Answer
-
The term is (62)24⋅x4⋅72y2, so the term is 15⋅16⋅49x4y2=11760x4y2
Exercise 7.6.8
What is the term in (x−3y)8 with x5?