Precalculus by Richard Wright

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10-06 Binomial Theorem

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.6.4, PC.7.2

Yanghui triangle
Figure 1: Yanghui's triangle. credit (wikimedia)

A binomial expression is two terms such as x + y. You already know how to work with binomials including how to calculate exponents of binomials by multiplying the binomial by itself. However, calculating higher exponents of binomials can be long and tedious. The binomial theorem is a shortcut to expand exponents of binomials.

The first 6 powers of (x + y)n are given in the triangle below. The plus signs + between the terms have been removed to simplify the diagram. The coefficients make a triangle called Pascal's Triangle. Blaise Pascal wrote a treatise on the triangle in 1654. However, it was first known in Persia by Al-Karaji and China by Yang Hui around 1000 AD. It is created by adding the two upper numbers to get the one below. Such as the 3 + 3 in the n = 3 row give the 6 in the n = 4 row.

$$ \begin{array}{lcr} (x + y)^0 & 1 & n = 0 \\ (x + y)^1 & 1x \quad 1y & n = 1 \\ (x + y)^2 & 1x^2 \quad 2xy \quad 1y^2 & n = 2 \\ (x + y)^3 & 1x^3 \quad 3x^2y \quad 3xy^2 \quad 1y^3 & n = 3 \\ (x + y)^4 & 1x^4 \quad 4x^3y \quad 6x^2y^2 \quad 4xy^3 \quad 1y^4 & n = 4 \\ (x + y)^5 & 1x^5 \quad 5x^4y \quad 10 x^3y^2 \quad 10 x^2y^3 \quad 5xy^4 \quad 1y^5 & n = 5 \\ & \nearrow \qquad \nearrow \qquad \nearrow \quad \qquad \nearrow \qquad \nearrow \qquad \nearrow \quad & \\ & r=0 \quad r=1 \quad r=2 \quad r=3 \quad r=4 \quad r=5 \quad & \end{array} $$

Pascal's Triangle creation
Figure 2: Create Pascal's Triangle by adding the two numbers above to get the one below. credit (wikimedia)

Notice there are some patterns in the triangle.

  1. Each row has n + 1 terms.
  2. In each row the powers of x count down and the powers of y count up.
  3. In each term, the sum of the exponents = n.
  4. The coefficients in each row are symmetrical. These are called combinations, \(_nC_r\).

The binomial coefficients, or combinations, are calculated by \(_nC_r = \frac{n!}{(n-r)!r!}\). This can also be represented \(\left(\begin{matrix} n \\ r \end{matrix}\right)\). Most calculators have a combination function built in.

Binomial Coefficient, or Combination

$$ _nC_r = \frac{n!}{(n-r)!r!} $$

Or another notation.

$$ \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{n!}{(n-r)!r!} $$

Combinations on TI-84
  1. Enter the value of n
  2. Press MATH
  3. Use the arrow pad to move to the PRB menu
  4. Go down to select the nCr
  5. Enter the value of r
  6. Press ENTER

Entered as

n nCr r

Combinations on NumWorks
  1. Select Calculation from the home screen
  2. Press the Toolbox button
  3. Go down and select the Probability menu
  4. Select Combinatorics
  5. Select \(\left(\begin{matrix}n \\ k\end{matrix}\right)\)
  6. Enter the value of n on top and the value of r on the bottom
  7. Press EXE

Entered as

\(\left(\begin{matrix}n \\ r\end{matrix}\right)\)

Example 1: Evaluate a Combination

Evaluate \(_6C_3\).

Solution

You can either use the formula or the button on the calculator. First the formula. \(n = 6, r = 3\)

$$ _nC_r = \frac{n!}{(n-r)!r!} $$

$$ _6C_3 = \frac{6!}{(6-3)!3!} = 20 $$

You might rather use the calculator function.

On a TI-84, type

6 nCr 3 ENTER

On a NumWorks, input

$$ \left(\begin{matrix} 6 \\ 3 \end{matrix}\right) $$

The output is 20.

Example 2: Evaluate Combinations

Evaluate (a) \(\left(\begin{matrix} 12 \\ 7 \end{matrix}\right)\) and (b) \(\left(\begin{matrix} 5 \\ 2 \end{matrix}\right)\)

Solutions
  1. $$ \left(\begin{matrix} 12 \\ 7 \end{matrix}\right) =\ _{12}C_7 = 792 $$
  2. $$ \left(\begin{matrix} 5 \\ 2 \end{matrix}\right) =\ _5C_2 = 10 $$
Try It 1

Evaluate (a) \(_{13}C_{10}\) and (b) \(\left(\begin{matrix} 8 \\ 1 \end{matrix}\right)\)

Answers

286; 8

Binomial Theorem

The binomial theorem is a shortcut for expanding powers of binomials.

Binomial Theorem

$$ (a + b)^n = \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$

  1. Plug in the a, b, and n. r starts at 0 and increases each term until r = n.
  2. Simplify each coefficient and exponent.
  3. Simplify each term.

Example 3: Expand a Binomial

Expand \((x + 3)^4\).

Solution

Compare \((x + 3)^4\) to \((a + b)^n\) to see that \(a = x, b = 3, n = 4\). Fill in the binomial theorem with r starting at 0.

$$ \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$

$$ \color{blue}{\ _4C_0 x^{4-0} 3^0} + \color{purple}{\ _4C_1 x^{4-1} 3^1} + \color{red}{\ _4C_2 x^{4-2} 3^2} + \color{orange}{\ _4C_3 x^{4-3} 3^3} + \color{green}{\ _4C_4 x^{4-4} 3^4} $$

$$ \color{blue}{1·x^4·1} + \color{purple}{4·x^3·3} + \color{red}{6·x^2·9} + \color{orange}{4·x^1·27} + \color{green}{1·x^0·81} $$

$$ \color{blue}{x^4} + \color{purple}{12x^3} + \color{red}{54x^2} + \color{orange}{108x} + \color{green}{81} $$

Example 4: Expand a binomial

Expand \((2 - x^3)^5\).

Solution

Compare \((2 - x^3)^5\) to \((a + b)^n\) to see that \(a = 2, b = -x^3, n = 5\). Fill in the binomial theorem with r starting at 0.

$$ \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$

$$ \color{blue}{\ _5C_0 2^{5-0} (-x^3)^0} + \color{purple}{\ _5C_1 2^{5-1} (-x^3)^1} + \color{red}{\ _5C_2 2^{5-2} (-x^3)^2} + \color{orange}{\ _5C_3 2^{5-3} (-x^3)^3} + \color{green}{\ _5C_4 2^{5-4} (-x^3)^4} + \color{brown}{\ _5C_5 2^{5-5} (-x^3)^5} $$

$$ \color{blue}{1·32·1} + \color{purple}{5·16·(-x^3)} + \color{red}{10·8·(x^6)} + \color{orange}{10·4·(-x^9)} + \color{green}{5·2·(x^{12})} + \color{brown}{1·1·(-x^{15})} $$

$$ \color{blue}{32} \color{purple}{- 80x^3} + \color{red}{80x^6} \color{orange}{- 40x^9} + \color{green}{10x^{12}} \color{brown}{- x^{15}} $$

Try It 2

Expand \((2x + 3)^3\).

Answer

\(8x^3 + 36x^2 + 54x + 27\)

Example 5: Find a Binomial Coefficient

Find the coefficient of the term \(x^5y^3\) in the expansion of \((3x - 2y)^8\).

Solution

Use the binomial theorem to find the coefficients. \(a = 3x, b = -2y, n = 8\) Because only one coefficient is needed, only one value of r is required. The binomial theorem for each term is

$$ _nC_r a^{n-r}\color{blue}{b^r} $$

Compare this to the desired term \(x^5\color{blue}y^3\), the exponent of the y should be the same as the exponent of the b part of the formula. Thus \(r=3\) and we only have to work through the one term.

$$ _8C_3 (3x)^{8-3} (-2y)^3 $$

$$ 56 (243x^5) (-8y^3) $$

$$-108864x^5y^3 $$

The coefficient of the \(x^5y^3\) is –108864.

Try It 3

Find the coefficient of the term \(x^4y^6\) in the expansion of \((x - 5y)^{10}\).

Answer

3281250

Lesson Summary

Binomial Coefficient, or Combination

$$ _nC_r = \frac{n!}{(n-r)!r!} $$

Or another notation.

$$ \left(\begin{matrix} n \\ r \end{matrix}\right) = \frac{n!}{(n-r)!r!} $$


Combinations on TI-84
  1. Enter the value of n
  2. Press MATH
  3. Use the arrow pad to move to the PRB menu
  4. Go down to select the nCr
  5. Enter the value of r
  6. Press ENTER

Entered as

n nCr r


Combinations on NumWorks
  1. Select Calculation from the home screen
  2. Press the Toolbox button
  3. Go down and select the Probability menu
  4. Select Combinatorics
  5. Select \(\left(\begin{matrix}n \\ k\end{matrix}\right)\)
  6. Enter the value of n on top and the value of r on the bottom
  7. Press EXE

Entered as

\(\left(\begin{matrix}n \\ r\end{matrix}\right)\)


Binomial Theorem

$$ (a + b)^n = \sum_{r=0}^n \ _nC_r a^{n-r} b^r $$

  1. Plug in the a, b, and n. r starts at 0 and increases each term until r = n.
  2. Simplify each coefficient and exponent.
  3. Simplify each term.

Helpful videos about this lesson.

Practice Exercises

    Evaluate the combination.

  1. \(_3C_2\)
  2. \(_6C_4\)
  3. \(\left(\begin{matrix} 11 \\ 4 \end{matrix}\right)\)
  4. \(\left(\begin{matrix} 15 \\ 7 \end{matrix}\right)\)
  5. \(\left(\begin{matrix} 7 \\ 7 \end{matrix}\right)\)
  6. Expand the binomial.

  7. \((x + 5)^3\)
  8. \((x - 3)^5\)
  9. \((2x + y)^4\)
  10. \((2x - 5y)^5\)
  11. \((3a + 7b)^6\)
  12. Find the specific coefficient of the binomial expansion.

  13. \(x^7\) term in \((x + 4)^{15}\)
  14. \(x^{13}y^7\) term in \((x + y)^{20}\)
  15. \(x^3y^{14}\) term in \((x - 2y)^{17}\)
  16. \(x^4y^6\) term in \((4x - 3y)^{10}\)
  17. Expand and simplify the difference quotient \(\displaystyle \frac{f(x+h)-f(x)}{h}\).

  18. \(f(x) = x^4\)
  19. Mixed Review

  20. (10-05) Prove \(5 + 7 + 9 + 11 + \cdots + (2n + 3) = n(n + 4)\).
  21. (10-05) Prove \(1 + 3 + 9 + 27 + \cdots + (3^{n-1}) = \frac{3^n-1}{2}\).
  22. (10-04) Write the rule for the nth term: 512, 256, 128, 64, ….
  23. (10-02) Evaluate \(\displaystyle \sum_{n=1}^{10} 2n\).
  24. (10-01) Write the first five terms of the sequence \(a_{n+1} = 2a_n - 3; a_1 = 4\).

Answers

  1. 3
  2. 15
  3. 330
  4. 6435
  5. 1
  6. \(x^3 + 15x^2 + 75x + 125\)
  7. \(x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243\)
  8. \(16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4\)
  9. \(32x^5 - 400x^4y + 2000x^3y^2 - 5000x^2y^3 + 6250xy^4 - 3125y^5\)
  10. \(729a^6 + 10206a^5b + 59535a^4b^2 + 185220a^3b^3 + 324135a^2b^4 + 302526ab^5 + 117649b^6\)
  11. 421724160
  12. 77520
  13. 11141120
  14. 39191040
  15. \(4x^3 + 6x^2h + 4xh^2 + h^3\)
  16. Show work
  17. Show work
  18. \(a_n = 512\left(1/2\right)^{n-1}\)
  19. 110
  20. 4, 5, 7, 11, 19