Week 3 – SS2 Third Term Further Mathematics Notes

WEEK THREE
TOPIC:BINOMIAL EXPANSION: PASCAL TRIANGLE, BINOMIAL THEOREM OF NEGATIVE, POSITVE AND FRACTIONAL POWER
PASCAL’S TRIANGLE
Consider the expressions of each of the following:
(x + y)⁰; (x + y )¹; (x + y)²; (x + y)³; (x + y)⁴
(x + y)⁰ = 1
(x + y)¹ = 1x + 1y
(x + y)² = 1x² + 2xy + 1y²
(x + y)³= 1x³ + 3x²y + 3xy² + 1y³
(x + y)⁴ = 1x⁴ + 4x³y + 6x²y² + 4xy³ + 1x⁴
The coefficient of x and y can be displayed in an array as:
                1
            1        1
        1        2        1
    1        3        3        1
1        4        6        4        1
The array of coefficients displayed above is called Pascal’s triangle, and it is used in determining the co-efficients of the terms of the powers of a binomial expression
Coefficient of (x + y)⁰                        1
Coefficient of (x + y)¹                    1        1
Coefficients of (x + y)²                1        2        1
Coefficients of (x + y)³            1        3        3        1
Coefficients of (x + y)⁴        1        4        6        4        1

 Example 1
Using Pascal’s riangle, expand and simplify completely: (2x + 3y)⁴
Solution:
(2x + 3y)⁴ = (2x)⁴ + 4(2x)³ (3y) + 6(2x)²(3y)² + 4(2x)(3y)³ + (3y)⁴
= 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴

 Examples 2:
Using pascal’s triangle, the coefficients of (x + y)⁵are: 1,5,10,10,5,1.
Therefore (x – 2y)⁵    = x⁵ + 5x⁴(-2y) + 10x³(-2y)² + 10x²(-2y)³ + 5x(-2y)⁴ + (-2y)⁵
            = x⁵ – 10x⁴y + 40x³y² – 80x²y³ + 80xy⁴ – 32y⁵
Example 3
Using Pascal’s triangle, simplify, correct to 5 decimal places (1.01)4
Solution
We can write (1.01)⁴ = (1 + 0.01)⁴
(1 + 0.01)⁴ = 1 + 4(0.01) + 6(0.01)² + 4(0.01)³+(0.01)⁴
= 1 + 0.04 + 0.0006 + 0.000004 + 0.00000001
= 1.04060401
= 1.04060 (5 d.p)
The Binomial Expansion Formula
Consider the expansion of (x + y)⁵ again
(x + y)⁵ = (x + y)(x + y)(x + y)(x + y)(x + y)
The first term is obtained by multiplying the xs in the five brackets. there is only one way to doing this
(x + y)ⁿ = xn + nx^{n – 1}y + xⁿ – ²y² + x^n-3y³ + ….
x^n-ry^r + …. yⁿ
It can be shown that the binomial expansion formula holds for positive, negative, integral or any rational value of n, provided there is a restriction on the values of x and y in the expansion of (x + y)ⁿ
We shall however consider only the binomial expansion formula for a positive integral n

 Example 4:

1. Write down the binomial expansion of ⁶ simplifying all the terms
2. Use the expansion in (a) to evaluate (1.0025)⁶ correct to five significant figures.

   Solution
   ⁶ = 1 + ⁶C₁ ¹ + ⁶C₂²
   + ⁶C₃ ³ + ⁶C₄
   + ⁶C₅ ⁵ + ⁶C₆ ^{6
   6}= 1 + x + ³ + x
   ⁴ + x ⁵ + ⁶
   = 1 + x + x²+ x³ +x⁴ + x⁵ + x⁶

3. (1.0025)⁶ = (1 + 0.0025)⁶

   = )⁶
   = )⁶
   Put x =
   x = x 4 = = 0.01
   therefore (1.0025)⁶ = 1 + (0.01) + (0.01)² + (0.01) +(0.01)⁴ + …
   = 1 + 0.015 + 0.00009375 + 0.0000003125
   = 1.0150940625
   = 1.0151 (5 s.f.)

    EVALUATION
   Expand ( 2 + 4x )⁴ simplifying the terms

    Example 5(a) Using the binomial theorem, obtain the expansion of (1 + 3x)⁶ + (1 – 3x)⁶ simplifying all the terms
   (b)Use the above result to calculate the value of (1.03)⁶ + (0.97)⁶, correct to five decimal places
   Solution:
   (1 + 3x)⁶ = 1 + ⁶C₁ (3x) + ⁶C₂ (3x)² + ⁶C₃ (3x)³ _ ⁶C₄ (3x)⁴ _ ⁶C₅ (3x)⁵ + ⁶C₆ (3x)⁶ ….. (1)
   (1 – 3x)⁶ = 1 – ⁶C₁ (3x) + ⁶C₂ (3x)² – ⁶C₃ (3x)³ + ⁶C₄ (3x)⁴ _ ⁶C₅ (3x)⁵ + ⁶C₆ (3x)⁶ ….. (2)
   Adding (1) and (2)
   (1 + 3x)⁶ +(1 – 3x)⁶ = 2 + 2 x ⁶C₂ (3x)² + 2 x ⁶C₄ (3x)⁴ + 2 x ⁶C₆ (3x)⁶
   = 2 + 2 x 9x² + 2 x x 81x⁴ + 2 x 729x⁶
   = 2 + 270x² + 2430x⁴ + 1458x⁶
   (1.03)⁶ = (1 + 0.03)⁶
   (0.97)⁶ = (1 – 0.03)⁶
   Put 1 + 0.03 = 1 + 3x
   Therefore 3x = 0.03
   Therefore x = 0.01
   Hence
   (1.03)⁶ + (0.97)⁶ = 2 + 270(0.01)² + 2430(0.01)⁴ + 1458(0.01)
   = 2 + 0.027 + 0.0000243 + 2.0270243
   = 2.02702 (5 d.p)

    Example 6
   

4. Using the binomial theorem, expamd (1 + 2x)⁵, simplifying all the terms
5. Use your expansion to calculate the value of 1.02⁵, correct to six significant figures

   If the first three terms of the expansion of (1 + px)ⁿ in ascending powers of x are 1 + 20x + 160x,
   Find the values of n and p
   Solution:

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

   = 1 + 5.(2x) + . 4x² + . 8x³ + . 16x + 32x⁵
   = 1 + 10x + 40x² + 80x³ + 80x⁴ + 32x⁵

7. (1.02) = (1 + 0.02)

   Put 1 + 0.02 = 1 + 2x
   Therefore 2x = 0.02
   x = 0.01
   Hence:
   (1.02)⁵ = 1 + 10(0.01) + 40(0.01)² + 80(0.01)³ + 80(0.01)⁴ + 32(0.01)⁵
   = 1 + 0.1 + 0.004 + 0.0008 + 0.00000008
   = 1.10408 (6.s.f.)
   6.3 The Binomial Theorem for any index
   The Binomial expansion formula is also applicable to any index n, where n can be positive or negative integer or even a fraction
   If /x/ 1, then:
   (1 + x)ⁿ = 1 + nx + + + x⁴ + … where n may be a negative integer or a fraction.

    Example 7
   Use the Binomial expansion formula to obtain the first five terms of the expansion of (1 + x)^-2
   Solution:
   (1 + x)^-2 = 1 + (-2) ( x) + ( ( x)² + ( x)³ + ( x)⁴ + ….
   = 1 – x + 3. ² – 4.³ + 5.⁴

    

8. (1 + px)ⁿ = 1 + 20x + 160x² + …

   (1 + px)ⁿ = 1 + ⁿc₁ (px) + ⁿc₁ (px)² + …
   = 1 + npx + p²x² …
   = 1 + 20x + 160x² + …
   By equating coefficients
   np = 20                 …     (1)
   p² = 160            …    (2)
   From (1) p =             …    (3)
   Therefore p² =         …    (4)
   Substituting (4) into (2)
   x = 160
   x 200 = 160
   There 200(n – 1) = 160n
   200n – 200 = 160n
   200n – 160n = 200
   40n = 200
   n = 5
   From (3)p = = 4
   Hence, n = 5, p = 4

    Example 8
   Obtain the first four terms of the explanation of (2 + x)⁸in ascending powers of x. hence, find the value of (2.005)⁸, correct to five significant figures.
   Solution:
   (2 + x)⁸= 2⁸(1+ x)⁸
   = 2⁸[1 +⁸C₁( x) + ⁸C₂ ( )² + ⁸C₃ ( )³ + … ]
   = 2⁸[1 +8( x) + ( )² + ( )³ + …]
   = 2⁸[1 + 2X + X²+ X³ + …]
   Write 2.0.005
   Put 2 + x = 2 + 0.005
   Therefore x = 0.005
   Therfore x = 0.005 x 2
   = 0.01
   Hence,
   (2.005)⁸ = 2⁸[1 +2(0.01) + (0.01)²+ (0.01)³ ]
   (2.005)⁸ = 2⁸ + 2⁹(0.01) + 2⁶.7(0.01)² + 2⁵ x 7(0.01)³ + …
   = 256 + 5.12 + 0.0448 + 0.000224
   = 261.165025
   = 261.17 (5 s.f.)

    GENERAL EVALUATION
   1) Write down and simplify all the terms of the binomial expansion of ( 1 – x )⁶ . Use the expansion to evaluate 0.997⁶ correct to 4 dp
   2) Write down the expansion of ( 1 + ¼ x ) ⁵ simplifying all its coefficients
   3) Use the binomial theorem to expand ( 2 – ¼ x)⁵ and simplify all the terms
   4) Deduce the expansion of ( 1 – x +x² )⁶ in ascending powers of x

    Reading Assignment
   New Further Maths Project 2 page 73 – 78

    WEEKEND ASSIGNMENT
   If the first three terms of the expansion of ( 1 + px )ⁿ in ascending powers of x are 1 + 20v + 160x find the value of
   1) n a) 2 b) 3 c) 4 d) 5
   2) p a) 2 b) 3 c) 4 d) 5
   3) In the expansion of ( 2x + 3y )⁴ what is the coefficient of y⁴ a) 16 b) 81 c) 216 d) 96
   4) How many terms are in the expansion of ( 1 – 4x ) ⁵ a) 3 b) 5 c) 6 d) 8
   5) What is the third term in the expansion of ( 1 – 3x )⁶ in ascending powers of x a) 18 b) -540 c) 135 d) 729

    THEORY
   1) Using binomial theorem, write down and simplify the first seven terms of the expansion of ( 1 + 2x )¹⁰ in ascending powers of x
   2) Expand ( 2 + x )⁵ ( 1 – 2x ) ⁶ as far as the term in x³ . Evaluate ( 1.999 )⁵ ( 1.002 )⁶