Question 1

The middle term in the expansion of (x + 2)⁶ is:

Accepted Answer

For the binomial expansion (x + 2)⁶, we have n = 6 (even).

When n is even, there is exactly one middle term at position (n/2 + 1).

Middle term position: r + 1 = 6/2 + 1 = 3 + 1 = 4
So r = 3.

The middle term is:
T₄ = C(6,3)·x⁶⁻³·2³
   = C(6,3)·x³·8
   = [6!/(3!·3!)]·x³·8
   = 20·x³·8
   = 160x³

Verification: C(6,3) = (6·5·4)/(3·2·1) = 120/6 = 20 ✓

Question 2

In the binomial expansion of (2x + 3)⁵, what is the coefficient of x³?

Accepted Answer

Using the Binomial Theorem: (a + b)ⁿ = Σ C(n,r) · aⁿ⁻ʳ · bʳ for r = 0 to n.

For (2x + 3)⁵, we have a = 2x, b = 3, n = 5.

The general term is: Tᵣ₊₁ = C(5,r) · (2x)⁵⁻ʳ · 3ʳ

For the term containing x³, we need: 5 - r = 3, so r = 2.

T₃ = C(5,2) · (2x)³ · 3²
   = [5!/(2!·3!)] · 8x³ · 9
   = 10 · 8 · 9 · x³
   = 720x³

Therefore, the coefficient of x³ is 720.

Question 3

The middle term in the binomial expansion of (x + 1/x)⁶ is:

Accepted Answer

For (x + 1/x)⁶, we have n = 6 (even), so there are 7 terms total (r = 0, 1, 2, ..., 6).

The middle term occurs at position (n/2 + 1) = (6/2 + 1) = 4th term, which corresponds to r = 3.

General term: Tᵣ₊₁ = C(6,r) · x⁶⁻ʳ · (1/x)ʳ = C(6,r) · x⁶⁻ʳ · x⁻ʳ = C(6,r) · x⁶⁻²ʳ

For r = 3:
T₄ = C(6,3) · x⁶⁻⁶ = C(6,3) · x⁰ = C(6,3) · 1

C(6,3) = 6!/(3!·3!) = (6·5·4)/(3·2·1) = 120/6 = 20

Therefore, the middle term is 20.

Question 4

If the coefficient of x⁴ in the expansion of (1 + x)ⁿ is 15, then n equals:

Accepted Answer

Using the Binomial Theorem: (1 + x)ⁿ = Σ C(n,r) · xʳ for r = 0 to n.

The term containing x⁴ is: T₅ = C(n,4) · x⁴

The coefficient of x⁴ is C(n,4) = 15.

We need to solve: C(n,4) = 15

C(n,4) = n!/(4!(n-4)!) = n(n-1)(n-2)(n-3)/(4·3·2·1) = n(n-1)(n-2)(n-3)/24 = 15

Therefore: n(n-1)(n-2)(n-3) = 360

Testing values:
- n = 5: 5·4·3·2 = 120 (too small)
- n = 6: 6·5·4·3 = 360 ✓

Verification: C(6,4) = 6!/(4!·2!) = (6·5)/(2·1) = 15 ✓

Therefore, n = 6.

Question 5

In the expansion of (2x - 3y)⁴, the term independent of x is:

Accepted Answer

Using the Binomial Theorem: (2x - 3y)⁴ = Σ C(4,r) · (2x)⁴⁻ʳ · (-3y)ʳ for r = 0 to 4.

General term: Tᵣ₊₁ = C(4,r) · (2x)⁴⁻ʳ · (-3y)ʳ = C(4,r) · 2⁴⁻ʳ · (-3)ʳ · x⁴⁻ʳ · yʳ

For a term independent of x, the exponent of x must be 0:
4 - r = 0 ⟹ r = 4

T₅ = C(4,4) · 2⁰ · (-3)⁴ · x⁰ · y⁴
   = 1 · 1 · 81 · 1 · y⁴
   = 81y⁴

Therefore, the term independent of x is 81y⁴.

Question 6

The sum of the coefficients in the expansion of (3x - 2)⁵ is:

Accepted Answer

To find the sum of all coefficients in a binomial expansion, substitute x = 1 into the expression.

For (3x - 2)⁵, substituting x = 1:
(3(1) - 2)⁵ = (3 - 2)⁵ = 1⁵ = 1

Alternatively, using the Binomial Theorem:
(3x - 2)⁵ = Σ C(5,r) · (3x)⁵⁻ʳ · (-2)ʳ for r = 0 to 5

The sum of coefficients is:
Σ C(5,r) · 3⁵⁻ʳ · (-2)ʳ = (3·1 - 2)⁵ = 1⁵ = 1

Therefore, the sum of the coefficients is 1.

Question 7

The coefficient of x⁵ in the expansion of (2 + x)⁸ is:

Accepted Answer

Using the Binomial Theorem: (a + b)ⁿ = Σ C(n,r) · aⁿ⁻ʳ · bʳ

For (2 + x)⁸, we have a = 2, b = x, n = 8.

The general term is: Tᵣ₊₁ = C(8,r) · 2⁸⁻ʳ · xʳ

For the coefficient of x⁵, we need r = 5:
T₆ = C(8,5) · 2⁸⁻⁵ · x⁵
   = C(8,5) · 2³ · x⁵

Calculate C(8,5):
C(8,5) = 8!/(5!·3!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56

Coefficient of x⁵ = 56 × 2³ = 56 × 8 = 448

Question 8

In the expansion of (1 + x)ⁿ, if the coefficient of x⁴ is 15, then n equals:

Accepted Answer

Using the Binomial Theorem: (1 + x)ⁿ = Σ C(n,r) · xʳ

The general term is: Tᵣ₊₁ = C(n,r) · xʳ

For the coefficient of x⁴, we have r = 4:
Coefficient of x⁴ = C(n,4) = 15

We need to find n such that C(n,4) = 15:
C(n,4) = n!/(4!(n-4)!) = n(n-1)(n-2)(n-3)/(4 × 3 × 2 × 1) = n(n-1)(n-2)(n-3)/24 = 15

Therefore: n(n-1)(n-2)(n-3) = 360

Test values:
- n = 6: 6 × 5 × 4 × 3 = 360 ✓

Verification: C(6,4) = 6!/(4!2!) = (6 × 5)/(2 × 1) = 15 ✓

Question 9

The middle term in the expansion of (x + 1/x)⁶ is:

Accepted Answer

For (x + 1/x)⁶, we have n = 6 (even).

When n is even, there is one middle term at position (n/2 + 1) = (6/2 + 1) = 4.
So the middle term is T₄ (the 4th term), which corresponds to r = 3 in Tᵣ₊₁.

General term: Tᵣ₊₁ = C(6,r) · (x)⁶⁻ʳ · (1/x)ʳ = C(6,r) · x⁶⁻ʳ · x⁻ʳ = C(6,r) · x⁶⁻²ʳ

For r = 3:
T₄ = C(6,3) · x⁶⁻⁶ = C(6,3) · x⁰ = C(6,3) · 1

Calculate C(6,3):
C(6,3) = 6!/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 120/6 = 20

Therefore, the middle term is 20.

Question 10

The sum of the binomial coefficients in the expansion of (2x + 3y)⁵ is:

Accepted Answer

The binomial expansion of (2x + 3y)⁵ is:
(2x + 3y)⁵ = Σ C(5,r) · (2x)⁵⁻ʳ · (3y)ʳ

The binomial coefficients are C(5,0), C(5,1), C(5,2), C(5,3), C(5,4), C(5,5).

To find the sum of binomial coefficients, we use the standard result:
Σ C(n,r) for r = 0 to n equals 2ⁿ

This can be verified by setting x = 1 and y = 1 in the binomial expansion:
(2·1 + 3·1)⁵ = (2 + 3)⁵ = 5⁵

But the sum of binomial coefficients alone (ignoring the constants 2 and 3) is:
C(5,0) + C(5,1) + C(5,2) + C(5,3) + C(5,4) + C(5,5) = 2⁵ = 32

Alternatively, set a = 1 and b = 1 in (a + b)ⁿ:
(1 + 1)⁵ = 2⁵ = 32

Question 11

If the coefficient of x² in the expansion of (1 + x)ⁿ is 10, then the coefficient of x³ in the same expansion is:

Accepted Answer

Using the Binomial Theorem: (1 + x)ⁿ = Σ C(n,r) · xʳ

The coefficient of x² is C(n,2) = 10.

Step 1: Find n from C(n,2) = 10:
C(n,2) = n(n-1)/2 = 10
n(n-1) = 20
n² - n - 20 = 0
(n - 5)(n + 4) = 0
n = 5 (since n must be positive)

Step 2: Verify: C(5,2) = 5 × 4 / 2 = 10 ✓

Step 3: Find the coefficient of x³:
Coefficient of x³ = C(5,3) = 5!/(3!2!) = (5 × 4)/(2 × 1) = 10

Question 12

The constant term in the expansion of (x² + 1/x)⁶ is:

Accepted Answer

Using the Binomial Theorem: (a + b)ⁿ = Σ C(n,r)·aⁿ⁻ʳ·bʳ

For (x² + 1/x)⁶, the general term is:
Tᵣ₊₁ = C(6,r)·(x²)⁶⁻ʳ·(1/x)ʳ
     = C(6,r)·x²⁽⁶⁻ʳ⁾·x⁻ʳ
     = C(6,r)·x¹²⁻²ʳ⁻ʳ
     = C(6,r)·x¹²⁻³ʳ

For the constant term, the power of x must be zero:
12 - 3r = 0
3r = 12
r = 4

The constant term is:
T₅ = C(6,4)·x⁰ = C(6,4) = 6!/(4!·2!) = (6·5)/(2·1) = 15

Question 13

If the coefficient of x⁴ in the expansion of (1 + x)ⁿ is 35, then n equals:

Accepted Answer

Using the Binomial Theorem: (1 + x)ⁿ = Σ C(n,r)·xʳ

The general term is: Tᵣ₊₁ = C(n,r)·xʳ

The coefficient of x⁴ is C(n,4).

Given: C(n,4) = 35

C(n,4) = n!/(4!(n-4)!) = n(n-1)(n-2)(n-3)/(4·3·2·1) = n(n-1)(n-2)(n-3)/24 = 35

Therefore: n(n-1)(n-2)(n-3) = 35 × 24 = 840

Testing values:
- If n = 7: 7·6·5·4 = 840 ✓

Verification: C(7,4) = 7!/(4!·3!) = (7·6·5)/(3·2·1) = 210/6 = 35 ✓

Therefore, n = 7

Question 14

In the binomial expansion of (3x - 2y)⁴, the sum of all coefficients is:

Accepted Answer

To find the sum of all coefficients in a binomial expansion, substitute x = 1 and y = 1 into the expression.

For (3x - 2y)⁴, substitute x = 1 and y = 1:
(3(1) - 2(1))⁴ = (3 - 2)⁴ = 1⁴ = 1

Alternatively, using the Binomial Theorem:
(3x - 2y)⁴ = Σ C(4,r)·(3x)⁴⁻ʳ·(-2y)ʳ

The general term is: Tᵣ₊₁ = C(4,r)·3⁴⁻ʳ·(-2)ʳ·x⁴⁻ʳ·yʳ

The coefficient of each term is: C(4,r)·3⁴⁻ʳ·(-2)ʳ

Sum of all coefficients = Σ C(4,r)·3⁴⁻ʳ·(-2)ʳ = (3 - 2)⁴ = 1⁴ = 1

Question 15

In the expansion of (x + 1/x)¹⁰, find the constant term (independent of x).

Accepted Answer

Using the Binomial Theorem: (a + b)ⁿ = Σ C(n,r)·aⁿ⁻ʳ·bʳ

Here, a = x, b = 1/x, n = 10.

General term: Tᵣ₊₁ = C(10,r)·x¹⁰⁻ʳ·(1/x)ʳ
                    = C(10,r)·x¹⁰⁻ʳ·x⁻ʳ
                    = C(10,r)·x⁽¹⁰⁻ʳ⁻ʳ⁾
                    = C(10,r)·x⁽¹⁰⁻²ʳ⁾

For the constant term, the power of x must be zero:
10 - 2r = 0
2r = 10
r = 5

Constant term = T₆ = C(10,5)·x⁰ = C(10,5)

C(10,5) = 10!/(5!·5!) = (10·9·8·7·6)/(5·4·3·2·1)
        = 30,240/120
        = 252

Question 16

The sum of the binomial coefficients in the expansion of (1 + x)ⁿ is 256. Find the value of n.

Accepted Answer

The sum of all binomial coefficients in the expansion of (1 + x)ⁿ is obtained by setting x = 1:

(1 + 1)ⁿ = C(n,0) + C(n,1) + C(n,2) + ... + C(n,n) = 2ⁿ

Given: Sum of binomial coefficients = 256

Therefore: 2ⁿ = 256

Since 256 = 2⁸, we have:
2ⁿ = 2⁸

Comparing exponents: n = 8

Question 17

In the binomial expansion of (2x + 3)^5, find the coefficient of x^3.

Accepted Answer

Using the Binomial Theorem: (a + b)^n = Σ C(n,r) · a^(n-r) · b^r, where r = 0 to n.

For (2x + 3)^5, the general term is:
T_(r+1) = C(5,r) · (2x)^(5-r) · 3^r

For the term containing x^3, we need:
5 - r = 3, so r = 2

T_3 = C(5,2) · (2x)^3 · 3^2
    = [5!/(2!·3!)] · 8x^3 · 9
    = 10 · 8 · 9 · x^3
    = 720x^3

Therefore, the coefficient of x^3 is 720.

Question 18

The term independent of x in the expansion of (x + 1/x²)^9 is:

Accepted Answer

Using the Binomial Theorem: (a + b)^n = Σ C(n,r) · a^(n-r) · b^r

For (x + 1/x²)^9, the general term is:
T_(r+1) = C(9,r) · x^(9-r) · (1/x²)^r
        = C(9,r) · x^(9-r) · x^(-2r)
        = C(9,r) · x^(9-r-2r)
        = C(9,r) · x^(9-3r)

For the term independent of x, the exponent of x must be zero:
9 - 3r = 0
3r = 9
r = 3

The independent term is:
T_4 = C(9,3) · x^0 = C(9,3)
    = 9!/(3!·6!)
    = (9·8·7)/(3·2·1)
    = 504/6
    = 84

Therefore, the term independent of x is 84.

Question 19

If the coefficient of x^7 in the expansion of (ax² + 1/bx)^10 is 120, find the value of ab.

Accepted Answer

Using the Binomial Theorem for (ax + 1/(bx))^10:

General term: T_(r+1) = C(10,r) · (ax)^(10-r) · (1/(bx))^r
                      = C(10,r) · a^(10-r) · x^(10-r) · b^(-r) · x^(-r)
                      = C(10,r) · a^(10-r) · b^(-r) · x^(10-2r)

For the coefficient of x^4:
10 - 2r = 4
2r = 6
r = 3

The coefficient of x^4 is:
C(10,3) · a^7 · b^(-3) = 120
120 · a^7/b^3 = 120
a^7/b^3 = 1

This means a^7 = b^3.

If a = 1 and b = 1, then a^7 = 1 and b^3 = 1, so the equation is satisfied.

Therefore, ab = 1 · 1 = 1.

Question 20

In the binomial expansion of (1 + x)^n, if the coefficients of the 5th and 6th terms are equal, find n.

Accepted Answer

In the expansion of (1 + x)^n, the general term is:
T_(r+1) = C(n,r) · x^r

The 5th term (r=4) has coefficient C(n,4).
The 6th term (r=5) has coefficient C(n,5).

Given that these coefficients are equal:
C(n,4) = C(n,5)

Using the property of binomial coefficients:
C(n,r) = C(n, n-r)

If C(n,4) = C(n,5), then either:
(1) 4 = 5 (impossible), or
(2) 4 = n - 5, which gives n = 9

Verification:
C(9,4) = 9!/(4!·5!) = (9·8·7·6)/(4·3·2·1) = 3024/24 = 126
C(9,5) = 9!/(5!·4!) = (9·8·7·6)/(4·3·2·1) = 3024/24 = 126 ✓

Therefore, n = 9.

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