24
THE BINOMIAL THEOREM
CONSIDER THE 4TH POWER of the binomial a + b:
(a + b)4 = (a + b)(a + b)(a + b)(a + b).
On actually multiplying out and collecting like terms, we would find
(a + b)4 = a4 + 4a3b + 6a²b² + 4ab3 + b4 . . . . . . . .(1)
Notice: The literal factors are all the combinations of a and b where the sum of the exponents is 4: a4, a³b, a²b², ab³, b4.
The degree of each term is 4.
Thus, to "expand" (a + b)5, we would anticipate the following terms, in which the sum of all the exponents is 5:
(a + b)5 = ? a5 + ? a4b + ? a³b² + ? a²b³ + ? ab4 + ? b5
The question is, What are the coefficients?
They are called the binomial coefficients. In the expansion of
(a + b)4, the binomial coefficients are 1 4 6 4 1; line (1) above. Note the symmetry.
The solution to the problem of the binomial coefficients without actually multiplying out, is called the binomial theorem. It gives the coefficients for the expansion of (a + b)n.
The symbol for a binomial coefficient is 
. The upper index n is the exponent of the expansion; the lower index k indicates which term.
For example, when n = 5, each term in the exansion of (a + b)5 will look like this:
a5 − kbk
k will successively take on the values 0 through 5.
(a + b)5 = 
a5 + 
a4b + 
a3b² + 
a2b3 + 
ab4 + 
b5
Notice: The lower index is equal to the exponent of b. The first term, then, has k = 0, because in the first term, b appears as b0, which is 1.
Now, what are these binomial coefficients, ?
The theorem states that the binomial coefficients are none other than the combinatorial numbers, nCk .
= nCk
| (a + b)5 | = | 5C0 a5 + 5C1 a4b + 5C2 a3b² + 5C3 a2b3 + 5C4 ab4 + 5C5 b5 |
| = | 1a5 +  a4b +  a3b² +  a²b3 +  ab4 +  b5 |
|
| = | a5 + 5a4b + 10a3b² + 10a²b3 + 5ab4 + b5 | |
The binomial coefficients here are 1 5 10 10 5 1. Note the symmetry.
Note also that the coefficient of the first term is 1, and the coefficient of the second term is the same as the exponent of (a + b), which here is 5.
Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:
What follows the summation sign is the general term. Each term in the sum will look like that -- the first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n − k) + k, always equals n.
Example 1.
a) The term a8b4 occurs in the expansion of what binomial?
Answer. (a + b)12. The sum of 8 + 4 is 12.
b) In that expansion, what number is the coefficient of a8b4?
Answer. It is the combinatorial number,
| 12C4 | = | 12· 11· 10· 9 1· 2· 3· 4 |
= 495 |
Note again that the lower index, 4, is the exponent of b.
This same number is also the coefficient of a4b8, since 12C8 = 12C4.
Example 2. Expand (a − b)5.
Solution. We found the binomial coefficients to be 1 5 10 10 5 1. The difference with (a − b) is that the signs of the terms will alternate:
(a − b)5 = a5 − 5a4b + 10a3b² − 10a²b3 + 5ab4 − b5.
For, a − b = a + (−b), therefore each term will have the form
a5 − k(−b)k.
When k is even, (−b)k will be positive. But when k is odd, (−b)k will be negative.
Each odd power of b will have a negative sign.
Example 3. In the expansion of (x − y)15, calculate the coefficients of x3y12 and x2y13.
Solution. The coefficient of x3y12 is positive, because the exponent of
y is even. That coefficient is 15C12. But 15C12 = 15C3, and so we have
| 15· 14· 13 1· 2· 3 |
= | 455. |
The coefficient of x2y13, on the other hand, is negative, because the exponent of y is odd. The coefficient is − 15C13 = − 15C2. We have
| − | 15· 14 1· 2 |
= −105 |
Example 4. Write the first four terms of (a + b)n. Do not use factorials.
Solution.
| (a + b)n | = | nC0 an + nC1 an − 1b + nC2 an − 2b² + nC3 an − 3b3 |
| = | an
+ nan − 1b
+  an − 2b²
+  an − 3b3
| |
Notice: Each coefficient contains the previous coefficient. The coefficient
contains the previous coefficient, n.
The next coefficient,
contains the previous one.
Each coefficient is the previous coefficient multiplied by the exponent of a, and divided by the next exponent of b.
The next coeffient would be this one
multiplied by (n − 3), and divided by 4:
| n(n − 1)(n − 2)(n − 3) 1· 2· 3· 4 |
Example 5. Use the binomial theorem to expand (a + b)8.
Solution. The expansion will begin :
(a + b)8 = a8 + 8a7b
The first coefficient is always 1. The second is equal to the exponent of the expansion, which here is 8.
Now, the next coefficient will be 8 times the exponent of a -- 7 -- divided by 2. It will be 56/2 = 28. We have
(a + b)8 = a8 + 8a7b + 28a6b2
The next coefficient is 28· 6 -- divided by 3:
28· 6 / 3 = 28· 2 = 56.
(a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3
The next coefficient is 56· 5 -- divided by 4:
56· 5 / 4 = 14· 5 = 70.
(a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4
And now we have come to the point of symmetry. For, the coefficient of a3b5 is equal to the coeffient of a5b3, which is 56. And so on for the remaining coefficients. Here is the complete expansion:
| (a + b)8 | = | a8 + 8a7b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + |
| 28a2b6 + 8ab7 + b8. | ||
Example 6. Write the 5th term in the expansion of (a + b)10.
Solution. In the 1st term, k = 0. In the 2nd term, k = 1. And so on.
The index k of each term
is one less than the ordinal number of the term.
Thus in the 5th term, k = 4. The exponent of b is 4. The 5th term is
| 10C4 | = | 10· 9· 8· 7 1· 2· 3· 4 |
a6b4 = 210 a6b4 |
Pascal's triangle
This triangular array is called Pascal's Triangle. Each row gives the binomial coefficients. That is, the row 1 2 1 are the coefficients of (a + b)². The next row, 1 3 3 1, are the coefficients of (a + b)3; and so on.
To construct the triangle, write 1, and below it write 1 1. Begin and end each successive row with 1. To construct the intervening numbers, add the two numbers immediately above.
Thus to construct the third row, begin it with 1, then add the two numbers immediately above: 1 + 1. Write 2. Finish the row with 1.
To construct the next row, begin it with 1, and add the two numbers immediately above: 1 + 2. Write 3. Again, add the two numbers immediately above: 2 + 1 = 3. Finish the row with 1.
Example 8. Expand (x − 1)6.
Solution. According to Pascal's triangle, the coefficients are
1 6 15 20 15 6 1.
In the binomial, x is "a", and −1 is "b". The signs will alternate:
| (x − 1)6 | = | x6 − 6x5· 1 + 15x4· 1² − 20x3· 13 + 15x²· 14 − 6x· 15 + 16 |
| = | x6 − 6x5 + 15x4 − 20x3 + 15x² − 6x + 1 | |
Example 9. Expand (x + 2)3.
Solution. The coefficients are 1 3 3 1. x is "a", and 2 is "b".
| (x + 2)3 | = | x3 + 3x²· 2 + 3x· 2² + 23 |
| = | x3 + 6x² + 12x + 8 | |
Problem 1. In the expansion of (a + b)n, each term has the form
an − kbk,
where k successively takes on the value 0, 1, 2, . . ., n.
is the symbol for the binomial coefficient.
The binomial theorem is the statement that = ?
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
The combinatorial number, nCk.
Problem 2. Use factorials to write the general term in the expansion of (a + b)n.
| n! (n − k)! k! |
an − kbk |
Problem 3.
a) Calculate the coefficient of a4b6 in the expansion of (a + b)10.
10C6 = 10C4 = 210
b) The coefficient of which other term is the same? a6b4
c) In the expansion of (a + b)n, the coefficient of an − kbk is the same as
c) the coefficient of which other term?
akbn − k
Problem 4. Calculate the coefficient of
a) x17y3 in the expansion of (x + y)20. 1140
b) x3y17 in the expansion of (x + y)20. 1140
c) x3y17 in the expansion of (x − y)20. −1140
d) x2y18 in the expansion of (x − y)20. 190
e) x5y5 in the expansion of (x − y)10. −252
f) x10 in the expansion of (x − 1)15. −3003
Problem 5. Write the first four terms of (x + h)n. Do not use factorials.
| (x + h)n | = | xn + nxn−1h + | n(n − 1) 1· 2 |
xn−2h2 + | n(n − 1)(n − 2) 1· 2· 3 |
xn−3h3 |
Problem 6. Compute the first four terms of each of the following.
a) (a + b)15 a15 + 15a14b + 105a13b² + 455a12b3
b) (x − 1)20 x20 − 20x19 + 190x18 − 1140x17
Problem 7. Consider the expansion of (x + b)30.
a) What is the exponent of b in the 1st term? 0
b) What is the exponent of b in the 3rd term? 2
c) In the 25th term? 24
d) In the kth term? k − 1
e) Write the fourth term, with its coefficient. 4,060x27b3
Problem 8. Calculate each of the following.
a) The third term of (a + b)11. 55a9b²
b) The fifth term of (x − y)7. 35x3y4
c) The tenth term of (x − 1)12. −220x3
| d) The fifteenth term of (1 + | 1 x |
)18. 3060x−14 |
| e) The fourth term of (x − | 1 x |
)10. −120x4 |
Problem 9. Use Pascal's triangle to expand the following.
a) (a + b)3 = a3 + 3a²b + 3ab² + b3
b) (a − b)3 = a3 − 3a²b + 3ab² − b3
c) (x + y)4 = x4 + 4x3y + 6x²y² + 4xy3 + y4
d) (x − y)4 = x4 − 4x3y + 6x²y² − 4xy3 + y4
e) (x − 1)5 = x5 − 5x4 + 10x3 − 10x² + 5x − 1
f) (x + 2)5 = x5 + 10x4 + 40x3 + 80x² + 80x + 32
g) (2x − 1)3 = 8x3 − 12x² + 6x − 1
h) (1 − xy)7
= 1 − 7xy + 21x²y² − 35x3y3 + 35x4y4 − 21x5y5 + 7x6y6 − x7y7
In the following Topic we will give a proof of the binomial theorem.
Next Topic: Multiplication of sums
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