High School

Homework for Section 13.5: The Binomial Theorem

**Multiple Choice:** Choose the one alternative that best completes the statement or answers the question.

1. Evaluate the expression \(\binom{9}{1}\)
- A) 1
- B) 110
- C) 27
- D) 9

2. Evaluate the expression \(\binom{2}{2}\)
- A) 7
- B) 1
- C) 21
- D) 2

3. Evaluate the expression \(\binom{10}{3}\)
- A) 504
- B) 252
- C) 126
- D) 30,240

**Expand the expression using the Binomial Theorem.**

4. \((x - 1)^6\)
- A) \(x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1\)
- B) \(x^6 + 6x^5 + 30x^4 + 120x^3 + 360x^2 + 720x + 720\)
- C) \(x^6 - 6x^5 + 30x^4 - 120x^3 + 360x^2 - 720x + 720\)
- D) \(x^6 - 6x^5 - 15x^4 - 20x^3 - 15x^2 - 6x - 6\)

5. \((x + 7)^5\)
- A) \(x^5 + 35x^4 + 490x^3 + 3430x^2 + 12,005x + 16,807\)
- B) \(x^5 + 35x^4 + 980x^3 + 6860x^2 + 12,005x + 16,807\)
- C) \(x^5 + 35x^4 + 490x^3 + 6860x^2 + 12,005x + 16,807\)
- D) \(x^5 + 35x^4 + 490x^3 + 3430x^2 + 12,005x + 7\)

6. \((4x + 3)^3\)
- A) \(64x^3 + 12x^2 + 729\)
- B) \(64x^3 + 144x^2 + 108 + 27\)
- C) \(16x^2 + 243 + 9\)
- D) \(64x^3 + 144x^2 + 144x + 27\)

**Use the Binomial Theorem to find the indicated coefficient or term.**

7. The coefficient of \(x\) in the expansion of \((2x + 3)^3\)
- A) 54
- B) 108
- C) 36
- D) 9

8. The coefficient of \(x\) in the expansion of \((3x + 5)^5\)
- A) 1875
- B) 15,625
- C) 5625
- D) 9375

9. The coefficient of \(x\) in the expansion of \((3x + 4)^6\)
- A) 11,664
- B) 23,328
- C) 19,440
- D) 34,560

Answer :

All the solutions by applying the combination formula and binomial formula,

1. (11 9) = 55

2. (7 2) = 21

3. (10 5) = 252

4. The expansion of (x - 1)⁶ = x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1

5. The expansion of (x + 7)⁵ = x⁵ + 35x⁴ + 490x³ + 3430x² + 112005x + 16807

6. The expansion of (4x + 3)³ = 64x³ + 144x² + 108x + 27

7. the coefficient of x is 54.

8. the coefficient of x is 9375.

9. the coefficient of x is 19440.

Now simplify each part as;

1. (11 9)

It is equivalent to

= (11 2)

Since (n r) = n!/ r! (n - r)!

Hence we get;

11! / 2! 9!

55

Hence option D is true.

2. (7 2)

= 7!/2! (7 - 2)!

= 7!/2! 5!

= 21

Hence option C is true.

3. (10 5)

= 10! / 5! (10 - 5)!

= 10!/5! 5!

= 252

Hence option B is true.

4. The expansion of (x - 1)⁶;

Using binomial expansion;

(x - y)ⁿ = xⁿ - ⁿC₁ xⁿ⁻¹ (y) + ⁿC₂ xⁿ⁻² y² - ......... + (- 1)ⁿ yⁿ

(x - 1)⁶ = x⁶ - 6x⁵ + 15x⁴ - 20x³ + 15x² - 6x + 1

5. The expansion of (x + 7)⁵;

Using binomial expansion;

(x - y)ⁿ = xⁿ - ⁿC₁ xⁿ⁻¹ (y) + ⁿC₂ xⁿ⁻² y² - ......... + (- 1)ⁿ yⁿ

(x + 7)⁵ = x⁵ + 35x⁴ + 490x³ + 3430x² + 112005x + 16807

6. The expansion of (4x + 3)³;

Using binomial expansion;

(x - y)ⁿ = xⁿ - ⁿC₁ xⁿ⁻¹ (y) + ⁿC₂ xⁿ⁻² y² - ......... + (- 1)ⁿ yⁿ

(4x + 3)³ = 64x³ + 144x² + 108x + 27

7. Since we know that;

[tex]T_ {r + 1} = ^n C_ r (x)^{n - r} a^r[/tex]

Hence we get;

[tex]T_ {r + 1} = ^3 C_ r (2x)^{3 - r} 3^r[/tex]

For the coefficient of x;

3 - r = 1

r = 3 - 1

r = 2

Hence the coefficient of x is,

[tex]T_ {2 + 1} = ^3 C_ 2 (2x)^{3 - 2} 3^2[/tex]

[tex]T_ {2 + 1} = 54x[/tex]

Hence, option A is true.,

8. Since we know that;

[tex]T_ {r + 1} = ^n C_ r (x)^{n - r} a^r[/tex]

Hence we get;

[tex]T_ {r + 1} = ^5 C_ r (3x)^{5 - r} 5^r[/tex]

For the coefficient of x;

5 - r = 1

5 - 1 = r

r = 4

Hence the coefficient of x is,

[tex]T_ {4 + 1} = ^5 C_ 4 (3x)^{5 - 4} 5^4[/tex]

[tex]T_ {4 + 1} = 9375x[/tex]

Hence, option D is true.

9. Since we know that;

[tex]T_ {r + 1} = ^n C_ r (x)^{n - r} a^r[/tex]

Hence we get;

[tex]T_ {r + 1} = ^6 C_ r (3x)^{6 - r} 4^r[/tex]

For the coefficient of x⁴;

6 - r = 4

r = 2

Hence ;

[tex]T_ {r + 1} = ^6 C_ 2 (3x)^{6 - 2} 4^2[/tex]

[tex]T_ {2 + 1} = 19440[/tex]

Hence option C is true.

To learn more about the combination visit:

brainly.com/question/28065038

#SPJ4

Other Questions