the binomial expansion

The Binomial Expansion

Introduction• You first met the Binomial Expansion in C2

• In this chapter you will have a brief reminder of expanding for positive integer powers

• We will also look at how to multiply out a bracket with a fractional or negative power

• We will also use partial fractions to allow the expansion of more complicated expressions

Teachings for Exercise 3A

The Binomial ExpansionYou need to be able to expand expressions of the form (1 + x)n where n is any real

number

3A

Find: (1+𝑥 )4

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !……+¿𝑛𝐶𝑟 𝑥𝑟 ¿

(1+𝑥 )4 1¿ +(4 )𝑥+4 (3) 𝑥2

2+(4 )(3)(2) 𝑥

3

6+(4 )(3)(2)(1) 𝑥4

24

1¿ +4 𝑥+6 𝑥2+4 𝑥3+𝑥4

Every term after this one will contain a (0) so can be ignored

The expansion is finite and exact

Always start by writing out the general form

Sub in:n = 4x = x

Work out each term separately and simplify


number

3A

Find: (1−2 𝑥 )3

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !……+¿𝑛𝐶𝑟 𝑥𝑟 ¿

(1−2 𝑥 )31¿ +(3)(−2𝑥)+3 (2)(−2𝑥 )2

2+(3)(2)(1)

(−2 𝑥)3

6

1¿ −6 𝑥+12 𝑥2−8𝑥3

Every term after this one will contain a (0) so can be ignored

The expansion is finite and exact

Always start by writing out the general form

Sub in:n = 3

x = -2xWork out each term separately and

simplifyIt is VERY important to put brackets

around the x parts


number

3A

Find:1

(1+𝑥)

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1+𝑥 )−11¿ +(−1)(𝑥)+(−1)(−2)(𝑥)2

2+(−1)(−2)(−3)

(𝑥 )3

6

1¿−𝑥+𝑥2−𝑥3

Rewrite this as a power of x first

Sub in:n = -1x = x


¿¿Write out the general form (it is very unlikely you will have to go beyond the first 4

terms)

With a negative power you will not get a (0) term

The expansion is infinite It can be used as an approximation for the

original term


number

3A

Find:√1−3 𝑥

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1−3 𝑥 )12 1¿+( 12 )(−3 𝑥)+( 12 )(− 12 ) (−3 𝑥)2

2+(12 )(− 12 )(− 32 ) (−3 𝑥)3

6

1¿− 32 𝑥−98 𝑥

2− 2716

𝑥3

Rewrite this as a power of x first

Sub in:n = 1/2x = -3x

Work out each term separately and simplify You should use your

calculator carefully

¿¿Write out the general form (it is very unlikely you will have to go beyond the first 4

terms)

With a fractional power you will not get a (0) term

The expansion is infinite It can be used as an approximation for the

original term


number

3A

Find the Binomial expansion of:¿(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥

2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1−𝑥 )13 1¿ +( 13 )(−𝑥)+( 13 )(− 23 ) (− 𝑥)2

2+( 13 )(− 23 )(− 53 )(−𝑥)3

6

1¿ − 13 𝑥−19 𝑥

2− 581

𝑥3

Sub in:n = 1/3x = -x


Write out the general formand state the values of x for which it is valid…

Imagine we substitute x = 2 into the expansion1¿− 23−

49−4081

1¿ −0.666−0.444−0.4938

The values fluctuate (easier to see as decimals)

The result is that the sequence will not converge and hence for x = 2, the expansion

is not valid


number

3A


2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1−𝑥 )13 1¿ +( 13 )(−𝑥)+( 13 )(− 23 ) (− 𝑥)2

2+( 13 )(− 23 )(− 53 )(−𝑥)3

6

1¿ − 13 𝑥−19 𝑥

2− 581

𝑥3




Imagine we substitute x = 0.5 into the expansion

1¿− 16−136−

5648

1¿ −0.166 27 −0.0077

The values continuously get smaller This means the sequence will converge (like an infinite series) and hence for x =

0.5, the sequence IS valid…


number

3A


2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1−𝑥 )13 1¿ +( 13 )(−𝑥)+( 13 )(− 23 ) (− 𝑥)2

2+( 13 )(− 23 )(− 53 )(−𝑥)3

6

1¿ − 13 𝑥−19 𝑥

2− 581

𝑥3




How do we work out for what set of values x is valid?The reason an expansion diverges or converges is down to the x

term…If the term is bigger than 1 or less than -1, squaring/cubing etc will accelerate the size of the term, diverging

the sequenceIf the term is between 1 and -1, squaring and cubing cause the terms to become increasingly small, to the

sum of the sequence will converge, and be valid

−1<− 𝑥<1 ¿−𝑥∨¿1¿ 𝑥∨¿1

Write using

ModulusThe expansion is valid when

the modulus value of x is less than 1


number

3A

Find the Binomial expansion of: 1¿¿

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1+4 𝑥 )− 21¿ +(−2 )(4 𝑥)+(−2 ) (−3 )(4 𝑥)2

2+(−2 ) (−3 ) (−4 )

(4 𝑥)3

6

1¿ −8𝑥+48 𝑥2−256 𝑥3

Sub in:n = -2x = 4x


Write out the general form:

and state the values of x for which it is valid…

¿¿

The ‘x’ term is 4x…

|4 𝑥|<1

|𝑥|< 14Divide by 4


number

3A

Find the Binomial expansion of:√1−2𝑥

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !

(1−2 𝑥 )12 1¿ +( 12 )(−2𝑥 )+( 12 )(− 12 ) (−2 𝑥)2

2+( 12 )(− 12 )(− 32 ) (−2 𝑥)3

6

1¿ −𝑥− 12 𝑥2− 12𝑥3

Sub in:n = 1/2x = -2x



and by using x = 0.01, find an estimate for √2

¿¿


number

3A

Find the Binomial expansion of:√1−2𝑥and by using x = 0.01, find an estimate for √2

√1−2𝑥¿1−𝑥−12 𝑥

2− 12 𝑥3

x = 0.01√0.98¿1−0.01−0.00005−0.0000005

√ 98100¿0.98994957√ 210 ¿0.9899495

7 √2¿9.899495√ 2¿1.414213571

Rewrite left using a fraction

Square root top and bottom separately

Multiply by 10

Divide by 7

Teachings for Exercise 3B

The Binomial ExpansionYou can use the expansion for (1 + x)n to expand (a + bx)n by taking out a as a

factor

3B

Find the first 4 terms in the Binomial expansion of:√ 4+𝑥¿¿¿ [4 (1+ 𝑥4 )]

12

¿

¿ 412(1+𝑥4 )

12

¿2(1+ 𝑥4 )12

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !


(1+ 𝑥4 )121¿ +( 12 )( 𝑥4 )+( 12 )(− 12 )

(𝑥4 )2

2+( 12 )(− 12 )(− 32 )

(𝑥4 )3

6

(1+ 𝑥4 )12 1¿ +1

8 𝑥− 1128 𝑥

2 +11024 𝑥

3

2(1+ 𝑥4 )12 2¿ +1

4 𝑥− 164 𝑥2+1512 𝑥

3

Take a factor 4 out of the brackets

Both parts in the square brackets are to the power 1/2

You can work out the part outside the bracket

Sub in:n = 1/2x = x/4Work out each term

carefully and simplify it

Remember we had a 2 outside the bracket

Multiply each term by 2

|𝑥4 |<1|𝑥|<4

Multiply by 4

The Binomial ExpansionYou can use the expansion for (1 + x)n to expand (a + bx)n by taking out a as a

factor

3B

Find the first 4 terms in the Binomial expansion of: 1¿¿¿¿

¿ [2(1+ 3 𝑥2 )]− 2

¿

¿2− 2(1+ 3 𝑥2 )−2

¿14 (1+ 3 𝑥2 )

−2

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !


(1+ 3 𝑥2 )−2

1¿ +(−2 )( 3 𝑥2 )+(−2 ) (−3 )( 3 𝑥2 )

2

2+(−2 ) (−3 ) (−4 )

( 3 𝑥2 )3

6

(1+ 3 𝑥2 )−2

1¿ −3 𝑥+274 𝑥2− 272 𝑥3

14 (1+ 3𝑥2 )

− 214¿ − 34 𝑥

+2716 𝑥2− 278 𝑥3

Take a factor 2 out of the brackets

Both parts in the square brackets are to the power -2

You can work out the part outside the bracket

Sub in:n = -2x = 3x/2

Work out each term carefully and simplify it

Remember we had a 1/4 outside the bracket

Divide each term by 4

|3 𝑥2 |<1|𝑥|< 23

Multiply by 2, divide by 3

Teachings for Exercise 3C


3C

You can use Partial fractions to simplify the expansions of more difficult expressions

Find the expansion of: up to and including the term in x34−5 𝑥

(1+𝑥)(2− 𝑥)

Express as Partial Fractions4−5 𝑥

(1+𝑥)(2− 𝑥)¿

𝐴(1+𝑥 )

+𝐵(2−𝑥)

¿𝐴 (2−𝑥 )+𝐵(1+𝑥)

(1+𝑥 )(2−𝑥)

¿ 𝐴 (2−𝑥 )+𝐵(1+𝑥 )4−5 𝑥¿3𝐵−6¿𝐵−2¿3 𝐴9¿ 𝐴3

4−5 𝑥(1+𝑥)(2− 𝑥)

¿3

(1+𝑥 )− 2

(2−𝑥 )

Cross-multiply and combine

The numerators must be equal

If x = 2

If x = -1

Express the original fraction as Partial Fractions, using A and B


3C



(1+𝑥)(2− 𝑥)4−5 𝑥

(1+𝑥)(2− 𝑥)¿

3(1+𝑥 )

− 2(2−𝑥 )

¿3¿−2¿Expand each term separately

3¿

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !


(1+𝑥 )−11¿ +(−1)(𝑥)+(−1)(−2)(𝑥)2

2+(−1)(−2)(−3)

(𝑥 )3

6

1¿−𝑥+𝑥2−𝑥33 (1+𝑥 )−1 3¿−3 𝑥+3 𝑥2−3 𝑥3

Both fractions can be rewritten

Sub in:x = xn = -1Work out each term carefully

Remember that this expansion is to be multiplied

by 3


3C



(1+𝑥)(2− 𝑥)4−5 𝑥

(1+𝑥)(2− 𝑥)¿

3(1+𝑥 )

− 2(2−𝑥 )


2¿


3 (1+𝑥 )−1 3¿−3 𝑥+3 𝑥2−3 𝑥3

2[2 (1− 𝑥2 )]−1

2[2− 1(1− 𝑥2 )− 1]

2[ 12 (1− 𝑥2 )− 1]

(1− 𝑥2 )

−1

Take a factor 2 out of the brackets (and keep the current 2 separate…)

Both parts in the square brackets are raised to -1

Work out 2-1

This is actually now cancelled by the 2 outside the square

bracket!


3C



(1+𝑥)(2− 𝑥)4−5 𝑥

(1+𝑥)(2− 𝑥)¿

3(1+𝑥 )

− 2(2−𝑥 )


2¿


3 (1+𝑥 )−1 3¿−3 𝑥+3 𝑥2−3 𝑥3

¿ (1− 𝑥2 )− 1

(1+𝑥 )𝑛 1¿ +𝑛𝑥+𝑛 (𝑛−1) 𝑥2

2 !+𝑛 (𝑛−1)(𝑛−2) 𝑥

3

3 !


(1− 𝑥2 )

−1

1¿ +(−1)(− 𝑥2 )+(−1)(−2)(− 𝑥2 )

2

2+(−1)(−2)(−3)

(− 𝑥2 )3

6

1¿ +𝑥2

+𝑥24

+𝑥38

Sub in:x = -x/2n = -1Work out each term carefully

(1− 𝑥2 )−1


3C



(1+𝑥)(2− 𝑥)4−5 𝑥

(1+𝑥)(2− 𝑥)¿

3(1+𝑥 )

− 2(2−𝑥 )

¿3¿−2¿


3 (1+𝑥 )−1 3¿−3 𝑥+3 𝑥2−3 𝑥3

1¿ +𝑥2

+𝑥24

+𝑥38(1− 𝑥2 )

−1

¿ (3−3 𝑥+3 𝑥2−3 𝑥3)−(1+ 𝑥2 +𝑥24

+𝑥38 )

¿2− 72 𝑥+114 𝑥2− 258 𝑥3

Replace each bracket with its expansion

Subtract the second from the first (be wary of double negatives in

some questions)

Summary• We have been reminded of the Binomial Expansion

• We have seen that when the power is a positive integer, the expansion is finite and exact

• With negative or fractional powers, the expansion is infinite

• We have seen how to decide what set of x-values the expansion is valid for

• We have also used partial fractions to break up more complex expansions

the binomial expansion

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validthe binomial expansionyou

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