Solutions — Pascal’s Triangle and the Binomial Theorem

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1. Q1 — Pascal’s Triangle Rows

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   Building from row 6 (1 6 15 20 15 6 1), add adjacent pairs:

   Row 7:

   1, (1+6)=7, (6+15)=21, (15+20)=35, (20+15)=35, (15+6)=21, (6+1)=7, 1

   1   7   21   35   35   21   7   1

   Row 8:

   1, 8, 28, 56, 70, 56, 28, 8, 1

   1   8   28   56   70   56   28   8   1

   Row sums: 2⁷ = 128 and 2⁸ = 256. ✓

2. Q2 — Pascal’s Rule

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   (a) C(7, 3) = (7 × 6 × 5)/(3!) = 210/6 = 35.

   C(7, 4) = C(7, 3) = 35 (by symmetry).

   C(7, 3) + C(7, 4) = 35 + 35 = 70.

   C(8, 4) = (8 × 7 × 6 × 5)/(4!) = 1680/24 = 70. ✓

   (b) By Pascal’s rule C(n, r) + C(n, r+1) = C(n+1, r+1), with n = 9, r = 3:

   C(10, 4) = C(9, 3) + C(9, 4) = 84 + 126 = 210

3. Q3 — Expand Using the Binomial Theorem

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   (a) (x + 1)⁴: row 4 coefficients are 1, 4, 6, 4, 1.

   = C(4,0)x⁴(1)⁰ + C(4,1)x³(1)¹ + C(4,2)x²(1)² + C(4,3)x(1)³ + C(4,4)(1)⁴

   = x⁴ + 4x³ + 6x² + 4x + 1

   (b) (a + b)³: row 3 coefficients are 1, 3, 3, 1.

   = C(3,0)a³ + C(3,1)a²b + C(3,2)ab² + C(3,3)b³

   = a³ + 3a²b + 3ab² + b³

4. Q4 — Expand (x + 2)⁴

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   General term: T_r+1 = C(4, r) x^4−r 2^r

   r	C(4,r)	2r	Term
   0	1	1	x4
   1	4	2	8x3
   2	6	4	24x2
   3	4	8	32x
   4	1	16	16

   (x + 2)⁴ = x⁴ + 8x³ + 24x² + 32x + 16

5. Q5 — Find a Specific Term

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   General term of (x + 2)⁵: T_r+1 = C(5, r) x^5−r 2^r

   We need the power of x to be 3: set 5 − r = 3, giving r = 2.

   T₃ = C(5, 2) × x³ × 2² = 10 × 4 × x³

   The term containing x³ is 40x³.

6. Q6 — Find the Constant Term

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   Write the expression as (x + 3x⁻¹)⁶.

   General term: T_r+1 = C(6, r) (x)^6−r (3x⁻¹)^r = C(6, r)  3^r  x^6−r−r = C(6, r)  3^r  x^6−2r

   For a constant term: 6 − 2r = 0 ⇒ r = 3.

   T₄ = C(6, 3) × 3³ × x⁰ = 20 × 27 = 540

7. Q7 — Sum of Coefficients

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   (a) The sum of the binomial coefficients in (1 + x)⁸ is found by setting x = 1:

   (1 + 1)⁸ = 2⁸ = 256

   That is, C(8,0) + C(8,1) + C(8,2) + … + C(8,8) = 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256. ✓

   (b) Row 5 of Pascal’s triangle: 1, 5, 10, 10, 5, 1.

   Sum = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵. ✓

8. Q8 — Expand (2x − 3)⁴

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   Write as (2x + (−3))⁴. General term: T_r+1 = C(4, r) (2x)^4−r (−3)^r

   r = 0: 1 × 16x⁴ × 1 = 16x⁴

   r = 1: 4 × 8x³ × (−3) = −96x³

   r = 2: 6 × 4x² × 9 = 216x²

   r = 3: 4 × 2x × (−27) = −216x

   r = 4: 1 × 1 × 81 = 81

   (2x − 3)⁴ = 16x⁴ − 96x³ + 216x² − 216x + 81

   Note: the alternating sign pattern arises from (−3)^r.

9. Q9 — Find the Coefficient of x⁴

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   General term of (3 + 2x)⁷: T_r+1 = C(7, r)  3^7−r  (2x)^r = C(7, r)  3^7−r  2^r  x^r

   For x⁴: set r = 4.

   T₅ = C(7, 4) × 3³ × 2⁴ × x⁴

   C(7, 4) = C(7, 3) = (7 × 6 × 5)/(3!) = 35

   = 35 × 27 × 16 = 35 × 432 = 15 120

   The coefficient of x⁴ is 15 120.

10. Q10 — Coefficient of x³ in a Product Expansion

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    Method 1 (Combine exponents):

    (1 + x)⁵(1 + x)³ = (1 + x)⁸

    Coefficient of x³ = C(8, 3) = (8 × 7 × 6)/(3!) = 336/6 = 56

    Method 2 (Cross-multiplication verification):

    Collect all pairs (x^j from first) × (x^3−j from second), j = 0, 1, 2, 3:

    C(5,0)×C(3,3) + C(5,1)×C(3,2) + C(5,2)×C(3,1) + C(5,3)×C(3,0)

    = 1×1 + 5×3 + 10×3 + 10×1 = 1 + 15 + 30 + 10 = 56. ✓