BINOMIAL THEOREM WORKSHEET

1. Find the expansion of (2x + 3)⁵.

2. Find the expansion of (3x - 2)⁵.

3. Evaluate 98⁴.

4. Find the middle term in the expansion of (x + y)⁶.

5. Find the middle term in the expansion of (x + y)⁷.

6. Find the coefficient of x⁶ in the expansion of

(3 + 2x)¹⁰

7. Find the coefficient of x³ in the expansion of

(2 - 3x)⁷

8. Find the coefficient of x¹⁵ in the expansion of 

(x² + 1/x³)¹⁰

Answers

1. Answer :

(2x + 3)⁵

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b¹ + ⁿC₂a^n-2b² + ⁿC₃a^n-3b³.............  ...................+ ⁿC_ra^n-rb^r + ........ ⁿC_nbⁿ

Substitute a = 2x and b = 3.

(2x + 3)ⁿ

= ⁵C₀(2x)⁵ + ⁵C₁(2x)^5-13¹ + ⁵C₂(2x)^5-23²

+ ⁵C₃(2x)^5-33³ + ⁵C₄(2x)^5-43⁴ + ⁵C₅(2x)^5-53⁵

= 1(2x)⁵ + 5(2x)⁴(3) + 10(2x)³(9)

+ 10(2x)²(27) + 5(2x)¹(81) + 1(2x)⁰(243)

= 32x⁵ + 15(16x⁴) + 90(8x³) + 270(4x²) + 405(2x) + 243

= 32x⁵ + 240x⁴ + 7208x³ + 1080x² + 810x + 243

2. Answer :

(3x - 2)⁵

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁a^n-1b¹ + ⁿC₂a^n-2b² + ⁿC₃a^n-3b³.............  ...................+ ⁿC_ra^n-rb^r + ........ ⁿC_nbⁿ

Substitute a = 3x and b = -2.

(3x - 2)ⁿ

= ⁵C₀(3x)⁵ + ⁵C₁(3x)^5-1(-2)¹ + ⁵C₂(3x)^5-2(-2)²

+ ⁵C₃(3x)^5-3(-2)³ + ⁵C₄(3x)^5-4(-2)⁴ + ⁵C₅(3x)^5-5(-2)⁵

= 1(3x)⁵ + 5(3x)⁴(-2) + 10(3x)³(4) 

+ 10(3x)²(-8) + 5(3x)¹(16) + 1(3x)⁰(-32)

= 243x⁵ - 10(81x⁴) + 40(27x³) - 80(9x²) + 80(3x) - 32

= 243x⁵ - 810x⁴ + 1080x³ - 720x² + 240x - 32

3. Answer :

98⁴ = (100 - 2)⁴

= ⁴C₀(100)⁴ - ⁴C₁100^4-12¹ + ⁴C₂100^4-22²

- ⁴C₃100^4-32³ + ⁴C₄100^4-42⁴

= 1(100)⁴ - 4(100³)(2) + 6(100²)(4) 

- 4(100)(8) + 1(100⁰)(16)

= 100000000 - 4(1000000)(2) + 6(10000)(4)

- 4(100)(8) + 16

= 100000000 - 8000000 + 240000 - 3200 + 16

= 92236816

4. Answer :

(x + y)⁶

Here, the exponent is 6.

Since we have to find the middle in the expansion, divide the exponent 6 by 2.

6/2 = 3

So, the middle term in the expansion (x + y)⁶ is the term containing x³y³.

That is, 

= ⁶C₃x³y³

= 20x³y³

5. Answer :

(x + y)⁷

Here, the exponent is 7.

Since we have to find the middle in the expansion, divide the exponent 7 by 2.

7/2 = 3.5

3.5 is the value between 3 and 4.

So, there are two middle terms containing x⁴y³ and x³y⁴.

So, the middle term in the expansion (x + y)6 is the term containing x³y³.

They are

= ⁷C₃x⁴y³  and ⁷C₄x³y⁴

= 35x⁴y³  and 35x³y⁴

6. Answer :

In the expansion of (3 + 2x)¹⁰, we will have x⁶ in the term containing (2x)⁶.

Comparing (a + b)¹⁰ and (3 + 2x)¹⁰,

a = 3  and  b = 2x

In the expansion of (a + b)¹⁰, the term containing b⁶ is

= ¹⁰C₆a⁴b⁶

Substitute a = 3 and b = 2x.

= ¹⁰C₆(3)⁴(2x)⁶

= ¹⁰C₄(81)(64x⁶)

= 210(81)(64x⁶)

= 1088640x⁶

So, the coefficient of x⁶ is 1088640.

7. Answer :

In the expansion of (2 - 3x)⁷, we will have x³ in the term containing (-3x)³.

Comparing (a + b)⁷ and (2 - 3x)⁷,

a = 2  and  b = -3x

In the expansion of (a + b)⁷, the term containing b³ is

= ⁷C₃a⁴b³

Substitute a = 2 and b = -3x.

= ⁷C₃(2)⁴(-3x)³

= 35(16)(-27x³)

= -15120x³

So, the coefficient of x³ is -15120.

8. Answer :

(x² + 1/x³)¹⁰

In the expansion of (x² + 1/x³)¹⁰, 

1^st term will contain :

= (x²)¹⁰(1/x³)⁰

= x²⁰(1)

2^nd term will contain :

= (x²)⁹(1/x³)¹

= (x¹⁸)(1/x³)

= (x¹⁸)(x^-3)

= x^18-3

= x¹⁵

In the expansion of (x² + 1/x³)¹⁰, the send term contains x¹⁵.

The second term in the expansion of (x² + 1/x³)¹⁰,

= ¹⁰C₁(x²)⁹(1/x³)¹

= 10(x¹⁸)(x^-3)

= 10x^{18 - 3}

= 10x¹⁵

So, the coefficient of x¹⁵ is 10.

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