1 Finding the Binomial Theorem

Have you heard of the Binomial Theorem? It’s got a fancy name, so it must be important, right? The Binomial Theorem provides a formula for the expansion of powers of a binomial expression like a + b. To get a feel for the need of the theorem, what the theorem might say, and why anyone should care about it, you can look at a few relatively small powers of a + b.

1. Expand each of the following binomial powers. Here’s something to think about as you work: How can you get one answer from a previous one?
   1. (a + b)²
   2. (a + b)³
   3. (a + b)⁴
   4. (a + b)⁵
      Aren’t you glad you weren’t asked to expand and simplify (a + b)⁶? Well, you will in a moment, but first, . . .
      
   5. What patterns do you notice in the coefficients of these polynomials?
2. Predict what the expansion of (a + b)⁶ will look like.
3. Use the expansion of (a + b)⁵ to expand (a + b)⁶ and check to see if your prediction was correct.

Enough suspense! Are you ready for a quick method that determines the expansion of (a + b)ⁿ for any positive integer n? You may have already guessed the punch line:

THE BINOMIAL THEOREM:

(a + b)ⁿ = a⁰bⁿ + a¹b^n-1 + + a^n-1b¹ + aⁿb⁰.
The notation represents the quantity . For example, = = = 10. On many graphing calculators, these coefficients are listed as nCr. In order to compute , type 8 nCr 3.
The binomial theorem allows you to expand powers of a binomial expression quickly, especially if you know the values of the binomial coefficients, . For example,

2 Applying the Binomial Theorem

Here are a few examples to illustrate how much time can be saved using the theorem instead of multiplying everything out by hand.

1. 1. Expand (a+b)⁶ using the binomial theorem, being sure you get the same answer as in problem 5.
   2. Expand (a + b)⁷ using the binomial theorem.
   3. Expand (x + 2y)⁵ using the binomial theorem.
   4. Expand (3x - 2y)⁷ using the binomial theorem.
2. 1. What is the coefficient of a¹⁶⁹b⁶ in the expansion of (a + b)¹⁷⁵?
   2. What value of k, other than 169, will make the equation = true?
      Think about these problems without actually determining the value of 175!. After all, it would take a very long time by hand and many computer algebra systems wouldn’t be able to do it.
      
   3. What other term in the expansion of (a + b)¹⁷⁵ has the same coefficient as a¹⁶⁹b⁶?

1 Finding the Binomial Theorem

Hint to problem 1. While you aren’t prohibited from using technology (in the form of a computer algebra system, for example), it’s really not necessary. Just carefully multiply everything out and simplify. On subsequent calculations, be sure to make use of previous results.

Hint to problem 2. Look back at the results of problem 1. Is there a pattern to the coefficients? Where have you seen the coefficients before?

Hint to problem 3. Multiply your expansion for (a + b)⁵ by (a + b), being sure to carefully use the distributive property.

2 Applying the Binomial Theorem

Hint to problems 1(a) and (b). Determine the necessary binomial coefficients using the formula.

Hint to problem 1(c). Use the expansion of (a + b)⁵ with appropriate substitutions for a and b.

Hint to problem 1(d). Use the expansion of (a + b)⁷ with appropriate substitutions for a and b.

Hint to problem 2(a). Think how you might compute without attempting to determine 175!. Are there some terms you can cancel from both the numerator and denominator?

Hint to problem 2(b). What does “look like?”

Hint to problem 2(c). Don’t try to expand (a + b)¹⁷⁵! Think about what the binomial theorem says about these coefficients.

1 Finding the Binomial Theorem

1. 1. (a + b)² = a² + 2ab + b²
   2. (a + b)³ = a³ + 3a²b + 3ab² + b³
   3. (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
   4. (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵
2. x⁶ + 6x⁵y + 15x⁴y² + 20x³y³ + 15x²y⁴ + 6xy⁵ + y⁶
3. x⁶ + 6x⁵y + 15x⁴y² + 20x³y³ + 15x²y⁴ + 6xy⁵ + y⁶

2 Finding the Binomial Theorem

1. 1. a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶
   2. a⁷ + 14a⁶b + 84a⁵b² + 280a⁴b³ + 560a³b⁴ + 672a²b⁵ + 448ab⁶ + 128b⁷
   3. x⁵ + 10x⁴y + 40x³y² + 80x²y³ + 80xy⁴ + 32y⁵
   4. 2187x⁷ - 10206x⁶y + 20412x⁵y² - 22680x⁴y³ + 15120x³y⁴ - 6048x²y⁵ + 1344xy⁶ - 128y⁷
2. 1. 36582584325
   2. 
   3. a⁶b¹⁶⁹

1 Finding the Binomial Theorem

1. 1. (a + b)² = a² + 2ab + b²
   2. (a + b)³ = a³ + 3a²b + 3ab² + b³
   3. (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
   4. (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵
   5. The coefficients in the expansion of (a + b)ⁿ seem to be the entries in row n of Pascal’s Triangle.

      Alternatively, the coefficient of each term (other than aⁿ and bⁿ) in the expansion of (a + b)ⁿ is the sum of two coefficients in the expansion of (a + b)^n-1. For example, the coefficient of a^kb^5-k (when 0 < k < 5) in the expansion of (a + b)⁵ is the sum of the coefficients of a^kb^4-k and a^k-1b^5-k in the expansion of (a + b)⁴.
      Specifically, the coefficient of a²b³ in the expansion of (a + b)⁵ is the sum of the coefficients of a²b² and a¹b³ in the expansion of (a + b)⁴. This is due to the fact that the term a²b³ in the expansion of (a + b)⁵ is obtained by multiplying 4ab³ by a and by multiplying 6a²b² by b and adding the results.
      

2. The coefficients correspond to those numbers in the 6^th row of Pascal’s triangle.
   Teachers note: This prediction is not obvious, especially if students are not familiar with Pascal’s Triangle. You might want to provide some additional guidance.
   Alternatively, the coefficient of a^kb^6-k (when 0 < k < 6) in the expansion of (a + b)⁶ is the sum of the coefficients of a^kb^5-k and a^k-1b^6-k in the expansion of (a + b)⁵.

   However you look at it, this results in the prediction that (a + b⁶ = a⁶ + 6a⁵b + 15a⁴b² + 20a³b³ + 15a²b⁴ + 6ab⁵ + b⁶.

3. We know from problem 1 that (a+b)⁵ = a⁵+5a⁴b+10a³b²+10a²b³+5ab⁴+b⁵. Therefore, we know that
   

2 Applying the Binomial Theorem

1. 1. (a + b)⁶ = a⁰b⁶ + a¹b⁵ + a²b⁴ + a³b³ + a⁴by² + a⁵b¹ + a⁶b⁰, which simplifies to
      
      
      
      
   2. (a + b)⁷ = a⁰b⁷ + a¹b⁶ + a²b⁵ + a³b⁴ + a⁴by³+a⁵b²+a⁶b¹+a⁰b⁷, which equals
      
      
      
      
   3. Letting a = x and b = 2y in the expansion of (a + b)⁵, we have
      
   4. Let a = 3x and b = -2y. Then
      
2. 1. The coefficient of a^kb^175-k in the expansion of (a + b)¹⁷⁵ is , so the coefficient of a¹⁶⁹b⁶ is
      
   2. = = = ,
   3. Since is the coefficient of a⁶b¹⁶⁹ in the expansion of (a + b)¹⁷⁵, we know that a¹⁶⁹b⁶ and a⁶b⁶⁹ have the same coefficients.