Binomial Expansion Questions and Answers - Form 3 Topical Mathematics

Questions

1.  
   1. Expand (1 – 3x)⁵
   2. se your expansion to estimate the value of 0.997 Correct to 4 d.p.
2. 1. Expand (5 + ^X/₂) upto the term in X³
   2. Use your expansion to estimate the value of correct to one decimal place
3. 1. Expand ( 3 + 2x)⁶ up to the fourth term
   2. Use your expansion to estimate:- (3√3)⁶
4. Two dice are thrown once and their sum noted. Find the probability that the sum is odd
5. Find the length PR in a triangle PQR having PQ = 12cm, QR = 8.4cm angle QPR = 35^o and angle PRQ = 75^o leaving your answer correct to decimal places
6. 1. Use binomial expansion to evaluate (2+³/_x)⁵ up to the fifth term
   2. By expressing 9.5 in the form (2 + ³/_x), use the expansion in (a) above to calculate (9.5)⁵ correct to 3 d.p
7. Use the expansion of (x – 0.2)⁵ to find the exact value of 9.8⁵
8. Solve for x in the equation;
   log (x + 24) = 2 log 3 + log (9 – 2x).
9. Expand (1 + ^x/₁₂) in ascending powers of x upto the fourth term.
   Use the four terms to evaluate (⁵/₄)⁶ to 4 decimal places.
10.  
    1. Expand and simplify the binominal expression ( 1 + ½ x)⁸
    2. Use the expansion up to the fourth term to evaluate (1.05)⁸ to 2 decimal places
11. Expand (3 + x)⁴ in ascending powers of x. Use the first three terms of the expansion to evaluate (3.02)⁴, correct to 3 decimal places

Answers

1. 1. 1⁵ + 5 (-3x)¹ + 10(-3x)² + 10 (-3x)³ + 5(-3x)⁴ + (3x)⁵
      1 – 15x + 90x² – 270x³ + 405x⁴ – 243x⁵
      1 – 15x + 90x² – 270x³ + 405x⁴ – 243x⁵
   2. 3x = 1 – 0.997
      x = 0.001
      = 1 – 15 (0.001) + 90 (0.001)2 -270 (0.001)3 + 405(0.001)4
      = 1 – 0.015 + 0.00009 – 0.00000027 + ……………..
      = 1 + 0.00009 – 0.015 – 0.00000027
      = 1.00009 – 0.01500027= 0.98508973
      = -0.9851 (4 d.p)
2. 1. 5 + x⁶ = 15625 + 3125X + 9375X² + 625X³ +……
            2                      3            4            2
   2. X = 1
      (11)⁶ = 15625 + 3125 + 9375 + 625
        2                         3          4        2
      15625 + 1041.667 + 2343.75 + 312.5
3. (√3 + 2x)⁶ = (√3)⁶ + 6 (√3)⁵ 2x + 15 (√3)⁴ (2x)² + 20 ( √3)³(2x)³
   = 27 + 108x√3 + 270x² + 480x³√3
   √3+ 2x = 3√3
   √(2x) + 2√3
   x = 3
   27 + 108√3√3 + 270√3² + 480 √3(3)³
   = 27 + 324 + 810 + 4320= 5481
4.  
   

   	1	2	3	4	5	6
   1  2  3  4  5  6	2  3  4  5  6  7	3  4  5  6  7  8	4  5  6  7  8  9	5  6  7  8  9  10	6  7  8  9  10  11	7  8  9  10  11  12

   P(Sum odd) = ¹⁸/₃₆ = ½
5. ∠PQR = 180 – (35^o + 75)
   = 70^o
   PR² = 122 + 8.42 – 2(12)(8.4) Cos 70^o
   PR = 145.61 = 12.07
6. 1. Terms; 2⁵, 2⁴(³/_x), 2³(³/_x)² , 2²(³/_x)³, 2³(³/_x)⁴
      Co eff  1, 5, 10, 10, 5
      (2 + ³/_x)⁵ = 25 + 5(2)⁴(³/_x) + (2)³(³/_x)² + 10(22) ³/_x)2 + 5(2)(³/_x)
      = 32 + 2140x-1 + 720x-2 + 1080x³ + 820x -4
   2. 9.5 = 2+ ³/_x
      ³/_x = 7.5
      x=³/_7.5 = 0.4
      (9.5)⁵= 32 + 240 + 720 + 1086 +  810  
                          0.4    (0.4)²  (0.4)³    (0.4)⁴
      = 53647.625(3d.p)
7. X⁵ – 5x⁴ (0.2) + 10x³ (0.20 – 10x² (0.2)³ + 5x (0.2)⁴ – (0.2)⁵
   X⁵ – 5x⁴ (²/₁₀)² + 10x³(²/₁₀)² – 10x²(²/₁₀)³ + 5x (²/₁₀)⁴ – (²/₁₀)⁵+ x⁵ – (⁴/₁₀)x³ – (⁸/₁₀₀)x² +5 x 16 – 2⁵/_10⁵
   X⁵ – x⁴x³ – ⁸/₁₀₀x² + 80x – 2⁵/10⁵
   90, 392, 079
8. Log (x +24) = log(x(9-2x)
   X + 24 = 81-18x
   X =3
9. 1 + x = 1 + x + 5x² + 5x³ 
        12          2     48     432
   (1 + ^x/₁₂)⁶ = 1¼
   ^x/₁₂ = ¼
   12
   x = 3
   ⁵/₄ = 1 + ³/₂ + ⁹/₄₈ + ²⁷/₄₃₂
   = 2.7500
10. 1. (1 +½ )8=1+ 8(½) + 28(½ x)² + 56(½ x)³ + 70(½ x)⁴ + 567(½ x)⁵ + 2(½ x )⁶ + 8(½ x)⁷ + (½x)⁸
       = 1+ 4x + 7x² + 7x³ + 4.375x⁴ + 1.75x⁵ + 0.4375x⁶ + 0.0625x⁷ + ¹/₂₅₆x⁸
    2. (1.05)⁸ = 1 + 4(0.1) + 7(0.1)² + 7(0.1)³
       =1+ 0.4 + 0.07 + 0.0074…
       = 1.48
11. 81 + 27x + 9x² + 3x³ + xy
    81 + 108x + 54x³ + x⁴
    81 + 108 (0.02) + 54 (0.02)³
    = 83.182