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

x^{7} + 7x^{6}y + 21x^{5}?y² + 35x^{4}y³ + 35x³y^{4 }+ 21 x²y^{5 }+ 7xy^{6} + y^{7}.

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

^{n}C_{r}x^{n-1}y^{r }.

The fifth term of ( x + y )^{7} = ^{7}C_{4}x^{7 – 4}y^{4} = __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 + ^{6}C_{1}(¼x) +^{6}C_{2}(¼x)² + ^{6}C_{3}(¼x)^{3} + ^{6}C_{4}(¼x)^{4} + ^{6}C_{5}(¼x)^{5} +(¼x)^{6 }=

1 + 6*¼X + 15*¹/_{16}X² + 20*¹/_{64}X³ + 15*¹/_{256}X⁴ + 6*¹/_{1024}X^{5} + ¹/_{4096}X^{6 } =

1 + ³/_{2}X + ^{15}/_{16}X² + ^{5}/_{16}X³ + ^{15}/_{256}X⁴ + ³/_{512}X^{5} + ¹/_{4096}X^{6 }

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

(1 – ¼ x)^{6 }, where x = 10^{-2} =

1 + ³/_{2}X + ^{15}/_{16}X² + ^{5}/_{16}X³ + ^{15}/_{256}X⁴ + ³/_{512}X^{5} + ¹/_{4096}X^{6 }=

1 + ³/_{2}*10^{-2} + ^{15}/_{16}*10^{-4} + ^{5}/_{16}X*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 + ^{6}C_{1}(3x) +^{6}C_{2}(3x)² + ^{6}C_{3}(3x)^{3} + ^{6}C_{4}(3x)^{4} + ^{6}C_{5}(3x)^{5} +(3x)^{6}…………………..(1)

( 1 – 3x )^{6 }= 1 – ^{6}C_{1}(3x) +^{6}C_{2}(3x)² – ^{6}C_{3}(3x)^{3} + ^{6}C_{4}(3x)^{4} – ^{6}C_{5}(3x)^{5} +(3x)^{6}…………………..(2)

(1) + (2)

1 + 1 + ^{6}C_{2}(3x)² + ^{6}C_{2}(3x)² + ^{6}C_{4}(3x)^{4} + ^{6}C_{4}(3x)^{4} + (3x)^{6 }+ (3x)^{6}

2[1 + ^{6}C_{2}(3x)² + ^{6}C_{4}(3x)^{4} + (3x)^{6}] =

2[1 + 15* 9x² + 15* 18x⁴ +729x^{6}] =

2[1 + 135x² + 1215x⁴ + 729x^{6}] =

__2 + 270x² + 2430x⁴ + 1456x ^{6}__

(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 + 1456x^{6}

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 1.02^{5}, 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 + ^{5}C_{1}(2x) + ^{5}C_{2}(2x)² + ^{5}C_{3}(2x)³ + ^{5}C_{4}(2x)⁴ + (2x)^{5 }=

1 + 5*2x + 10*4x² + 10*8x³ + 5*16x⁴ + 32x^{5 }=

__1 + 10x + 40x² + 80x³ + 80x⁴ + 32x ^{5}__

(II) 1.02^{5 }= (1 + 0.02)^{5} = (1 + 2*10^{-2})^{5}

Let 10^{-2}= x.

This means that 1.02^{5 }= (1 + 2x)^{5}=

1 + 10x + 40x² + 80x³ + 80x⁴ + 32x^{5}=

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} = ^{n}C_{1}px + ^{n}C_{2}(px)² + …

^{n}C_{1}px = 20x, npx = 20x ,

np = 20…………………………….(1)

^{n}C_{2}(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

- 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 1.005^{6}, correct to five decimal places.

(b) Given that 3.15^{6} = 3^{6}*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 1.004^{8} 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 x^{5} (b) By taking x = 0.1, evaluate 0.91^{6}, 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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