Learning Resources
Lesson
We now want to extend what we discovered about the coefficients of (x + 1)n to the expansion of all binomials. We will look at (x + y)n and then use what we discover to expand binomials like (2x + 3y)5. However, let's first review what we discovered about (x + 1)n as it relates to Pascal's Triangle and extend that to combinations.
We learned that the coefficients of the expanded form of (x + 1)n are the elements of row n of Pascal's Triangle. In an earlier lesson we learned that the elements in row n of Pascal's Triangle are the combinations nC0 , nC1,nC2 , nC3 , nC4 , etc. Thus we can write:
(x + 1)n =nC0 xn + nC1 xn-1 + nC2 xn-2 + ...+ nCn-2 x2 + nCn-1 x + nCn 1
The above expression looks more complicated than it is. When it is written for a specific value of n it becomes more familiar. For example, if n = 4 we have:
(x + 1)4 =4C0 x4 + 4C1 x3 + 4C2 x2 + 4C3x + 4C41
(recall that nC0 and nCn are both equal to 1).
The coefficients of the expansion of (x + y)n have exactly the same relation to the elements of Pascal's Triangle as does the coefficients of the expansion of
(x + 1)n . However, we need to take a look at the exponents of the variables x and y in each of the terms of the expanded version. This is best seen by working through a few examples.
Note the exponents on the two variables. The exponent for the first variable starts at n, the exponent to which the binomial is raised, and decreases in each successive term by 1 till it reaches 0. The exponent for the second variable starts at zero (y0 = 1) in the first term and increases by 1 in each successive term till it reaches n.
Note also the coefficients of each of the terms in the expanded form are the elements in the corresponding row of Pascal's Triangle which can be found using combinations. Thus the coefficients may be written using combination notation similar to what was used above for (x + 1)n . Let's apply this to the expansion of a binomial to a higher power.
Example
Write the first 4 terms of the expansion of (x + y)9 .
Solution
We know that the coefficient of the first term of the expansion is 1 and the coefficient of the second term is 9. The coefficient of the third term is 9C2 and of the fourth term is 9C3. These are evaluated below:
It is now a matter of using the pattern for the exponents of the variables to write the first four terms of the expanded form.
(x + y)9 = 1x9 + 9x8y +36x7y2 + 84x6y3 + ........
Notice that the exponent of x starts at 9 and decreases while the exponent of y starts at 0 (and hence is not written down) and increases.
We now want to replace x and y in the binomial by other terms. The pattern of exponents and coefficients remain the same. We just have to be careful what term or expression we are raising to what power.
Example
Use your knowledge of Pascal's Triangle and combinations to expand the binomial (2a + 3b)6.
Solution
Think of (2a + 3b)6 as (x + y)6 with x = 2a and y = 3b. The expansion, written using combination notation is:
(2a + 3b)6 = 6C0(2a)6 + 6C1(2a)5(3b) + 6C2(2a)4(3b)2 + 6C3(2a)3(3b)3 +
6C4(2a)2(3b)4 + 6C5(2a)(3b)5 + 6C6(3b)6
If you have a copy of Pascal's Triangle to row 6 recorded in you note book, these values could have been read directly from there without having to do the calculations. Either way we have:
(2a + 3b)6 = 1(2a)6 + 6(2a)5(3b) + 15(2a)4(3b)2 + 20(2a)3(3b)3 +
15(2a)2(3b)4 + 6(2a)(3b)5 + 1(3b)6
Now simplify each of the terms (e.g. (2a)6 = (2a)(2a)(2a)(2a)(2a)(2a) = 64a6. This gives:
(2a + 3b)6 = 64a6 + 576a5b + 2160a4b2 + 4320a3b3 + 4860a2b4 + 2916ab5 + 729b6
Activity
- Read through Example 10 in Focus I on page 344 in your text.
- Complete the Focus Questions 15 & 16 on page 344.
- Do the CYU Questions 17 - 22 on page 345.
When you have completed these questions, ask your on-site teacher to get the solutions for you from the Teacher's Resource Binder and check them against your answers. After you do this, if there is something you had trouble with and still do not understand, contact your on-line teacher for help.
Test Yourself
Use combinations to expand the expression (3x - 2)5.