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

nCrxn-1yr .

The fifth term of ( x + y )7 = 7C4x7 – 4y4 = 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 + ¹/4096X=

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)46C5(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

(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.

1. Aji