# 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.
1. Expand (5 + X/2) upto the term in X3
2. Use your expansion to estimate the value of correct to one decimal place
1. Expand ( 3 + 2x)6 up to the fourth term
2. Use your expansion to estimate:- (3√3)6
2. Two dice are thrown once and their sum noted. Find the probability that the sum is odd
3. Find the length PR in a triangle PQR having PQ = 12cm, QR = 8.4cm angle QPR = 35o and angle PRQ = 75o leaving your answer correct to decimal places
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
4. Use the expansion of (x – 0.2)5 to find the exact value of 9.85
5. Solve for x in the equation;
log (x + 24) = 2 log 3 + log (9 – 2x).
6. Expand (1 + x/12) in ascending powers of x upto the fourth term.
Use the four terms to evaluate (5/4)to 4 decimal places.
7.
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
8. 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

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)
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
1. (√3 + 2x)6 = (√3)6 + 6 (√3)5 2x + 15 (√3)4 (2x)2 + 20 ( √3)3(2x)3
= 27 + 108x√3 + 270x2 + 480x33
√3+ 2x = 3√3
√(2x) + 2√3
x = 3
27 + 108√3√3 + 270√32 + 480 √3(3)3
= 27 + 324 + 810 + 4320= 5481
2.
 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
3. ∠PQR = 180 – (35o + 75)
= 70o
PR2 = 122 + 8.42 – 2(12)(8.4) Cos 70o
PR = 145.61 = 12.07
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)
4. 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
5. Log (x +24) = log(x(9-2x)
X + 24 = 81-18x
X =3
6. 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
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
7. 81 + 27x + 9x2 + 3x3 + xy
81 + 108x + 54x3 + x4
81 + 108 (0.02) + 54 (0.02)3
= 83.182

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