SS 2 FURTHER MATHS

FURTHER MATHEMATICS SSS2

BINOMIAL THEORY

The Binomial Expansion Formula

The Pascal’s triangle made binomial expression easier. Consider the expression ( x + y )⁷ . Rather than expanding this directly; (x + y)(x + y)(x + y)(x + y)(x + y)(x + y)(x + y), then continue with the step by step multiplication which is likely to consume up to 45minutes, it can be expanded using Pascal’s Triangle.

11

1 2 1

1 3 3 1

1 4  6 4  1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

x⁷ + 7x⁶y + 21x⁵?y² + 35x⁴y³ + 35x³y⁴ + 21 x²y⁵ + 7xy⁶ + y⁷.

However, the Binomial expansion formula makes it lots quite easier. In some cases just one or two terms of a binomial expression may be required. This formula makes it faster and easier to find just the required term(s).

The formula states that: the ( r + 1)th term of the expression ( x + y )ⁿ is given by

ⁿC_rx^n-1y^r .

The fifth term of ( x + y )⁷ = ⁷C₄x^{7 – 4}y⁴ = 35x³y⁴

Example 1

(a) Write down the Binomial expansion of         (1 – ¼ x)⁶, simplifying all terms

(b) Use the Binomial expansion to evaluate (1.0025)⁶, correct to five significant figures.

Solution

(a) (1 – ¼ x)⁶ = 1 + ⁶C₁(¼x) +⁶C₂(¼x)² + ⁶C₃(¼x)³ + ⁶C₄(¼x)⁴ + ⁶C₅(¼x)⁵ +(¼x)⁶ =

1 + 6*¼X + 15*¹/₁₆X² + 20*¹/₆₄X³ + 15*¹/₂₅₆X⁴ + 6*¹/₁₀₂₄X⁵ + ¹/₄₀₉₆X⁶  =

1 + ³/₂X + ¹⁵/₁₆X² + ⁵/₁₆X³ + ¹⁵/₂₅₆X⁴ +  ³/₅₁₂X⁵ + ¹/₄₀₉₆X⁶

(b) (1.0025)⁶ = (1 +0.0025)⁶ = (1 + ¼*10^-2)⁶=

(1 – ¼ x)⁶ , where x = 10^-2 =

1 + ³/₂X + ¹⁵/₁₆X² + ⁵/₁₆X³ + ¹⁵/₂₅₆X⁴ +  ³/₅₁₂X⁵ + ¹/₄₀₉₆X⁶  =

1 + ³/₂*10^-2 + ¹⁵/₁₆*10^-4 + ⁵/₁₆X*10^-6+ ¹⁵/₂₅₆*10^-8 +  ³/₅₁₂*10^-10 + ¹/₄₀₉₆*10^-12 =

1 + 0.015 + 0.00009375 +… = 1.01509375…

= 1.0151 to 5 SF

Example 2

(a) Using the Binomial theory obtain the expansion of ( 1 +3x )⁶ + ( 1 – 3x )⁶

(b) Use the above result to obtain the value of  ( 1.0 3)⁶ + ( 0.97)⁶ correct to five decimal places.

Solution

( 1 +3x )⁶ + ( 1 – 3x )⁶

( 1 +3x )⁶  = 1 + ⁶C₁(3x) +⁶C₂(3x)² + ⁶C₃(3x)³ + ⁶C₄(3x)⁴ + ⁶C₅(3x)⁵ +(3x)⁶…………………..(1)

( 1 – 3x )⁶ = 1 – ⁶C₁(3x) +⁶C₂(3x)² – ⁶C₃(3x)³ + ⁶C₄(3x)⁴ – ⁶C₅(3x)⁵ +(3x)⁶…………………..(2)

(1) + (2)

1 + 1 + ⁶C₂(3x)² + ⁶C₂(3x)²  + ⁶C₄(3x)⁴ + ⁶C₄(3x)⁴ + (3x)⁶ + (3x)⁶

2[1 + ⁶C₂(3x)² + ⁶C₄(3x)⁴ + (3x)⁶] =

2[1 + 15* 9x² + 15* 18x⁴ +729x⁶] =

2[1 + 135x² + 1215x⁴ + 729x⁶] =

2 + 270x² + 2430x⁴ + 1456x⁶

(b)  ( 1.0 3)⁶ = ( 1 +0.003)⁶ = ( 1 + 3*10^-2)⁶  = ( 1 + 3x )⁶. This means x = 10^-2

( 0.97)⁶ = ( 1 – 0.003)⁶ = ( 1 – 3*10^-2)⁶  = ( 1 – 3x )⁶. This means x = 10^-2

( 1.0 3)⁶ + ( 0.97)⁶ =

2 + 270x² + 2430x⁴ + 1456x⁶

2 + 270(10^-2)² + 2430(10^-2)⁴ + 1456(10^-2)⁶=

2 + 0.027 + 0.0000247… = 2. 0270243…

= 2.02702 to five decimal places.

Example 3

(a)(I) Using the Binomial theory expand            (1 + 2x)⁵, simplifying all terms.

(II) Use your expansion to calculate the value of 1.02⁵, correct to six significant figures

(b) If the first three terms of the expression  (1 + px)ⁿ in the ascending power of x is 1 + 120x + 160x², find the values of n and p.

Solution

(a)(I) (1 + 2x)⁵ = 1 + ⁵C₁(2x) + ⁵C₂(2x)² + ⁵C₃(2x)³ + ⁵C₄(2x)⁴ + (2x)⁵ =

1 + 5*2x + 10*4x² + 10*8x³ + 5*16x⁴ + 32x⁵ =

1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵

(II)  1.02⁵ = (1 + 0.02)⁵ = (1 + 2*10^-2)⁵

Let 10^-2= x.

This means that 1.02⁵ = (1 + 2x)⁵=

1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵=

1 + 10(10^-2) + 40(10^-2)² + 80(10^-2)³ + 80(10^-2)⁴ + 32(10^-2)⁵=

1 + 0.1 + 0.004 + 0.00008 + …

1.10408 to six significant figures

(b) (1 + px)ⁿ = 1 + 20x + 160x² …

(1 + px)ⁿ = ⁿC₁px + ⁿC₂(px)² + …

ⁿC₁px = 20x, npx = 20x ,

np = 20…………………………….(1)

ⁿC₂(px)²  = 1600x²

{ n! ÷ [( n- 2)!2!]}p²x² = 160x²

[ n! ÷ ( n – 2)!] p²x² = 2!*160x²

But [ n! ÷ ( n – 2)!] = n(n -1), and 2! =2

n(n – 1) p²x² = 2*160x²

np²(n – 1) = 320

npp(n – 1) = 230

But np = 20 (from equation 1)

20p(n – 1) = 320

p(n – 1) = 16

From equation 1: np = 20.

Therefore p = ²⁰/_n

²⁰/_n(n – 1) = 16

20(n – 1) = 16n

20n – 20 = 16n

20n – 16n = 20

4n = 20

n = ²⁰/₄

n = 5

p = ²⁰/_n

p = ²⁰/₅

p = 4

n = 5, p =4

RESEARCH WORK

1. Obtain the first four terms of (2 + ½X)⁸ in ascending power of X. Hence find the value of (2.005)⁸, correct to five significant figures.

2.(a)(I) Using the Binomial theory write down and simplify all terms of the expansion of (1 + ½X)⁶

(II)Use your answer to evaluate 1.005⁶, correct to five decimal places.

(b) Given that 3.15⁶ = 3⁶*b, use your answer to estimate the value of b, correct to four decimal places.

3.(a) Without using table find the value of (2 + √3 )⁶ + (2 – √3 )⁶ +

(b) Use binomial theory to evaluate 1.004⁸ correct to five decimal places.

4.(a) Expand (2 + x)⁵(1 – 2x)⁶ as far as the term in x³.

(b) By putting x = – 0.001 in your result, evaluate (1.999)⁵(1.002)⁶ , correct to five decimal places.

5(a)(I) Write down the binomial expansion of (1 – y)⁶ and simplify all terms.

(II) By putting y = x – x² in your answer, deduce the expansion of  (1 – x + x²)⁶ in ascending power of x as far as the term of x⁵ (b) By taking x = 0.1, evaluate 0.91⁶, correct to two decimal places.

ASSIGNMENT 1

This is on New Further Mathematics Project 2 by M. R Tuttuh-Adegun and D. God’spower Adegoke. 5th Edition. From page 85 and 86.

Exercise 6, from No1 to No10.

ASSIGNMENT 2

This is on New Further Mathematics Project 2 by M. R Tuttuh-Adegun and D. God’spower Adegoke. 5th Edition. From page 86 and 87.

Exercise 6, No 11, 12, 13, 15, 17, and 20.

NOTE: All works are to be submitted to this number; 08021443714 via WhatsApp.

DIAMOND ICE.

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