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Binomial expansions

Geometric sequences

Geometric series

The sum to infinity of a geometric series

Binomial expansions

Examination-style questions

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Pascal’s Triangle

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Binomial expansions

An expression containing two terms, for example a + b, is called a binomial expression.

When we find powers of binomial expressions an interesting pattern emerges.

(a + b)0 = 1

(a + b)1 = 1a + 1b

(a + b)2 = 1a2 + 2ab + 1b2

(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3

(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4

(a + b)5 = 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5

What patterns do you notice?

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Binomial expansions

In general, in the expansion of (a + b)n:

The coefficients are given by the (n + 1)th row of Pascal’s triangle.

The sum of the powers of a and b is n for each term.

Altogether, there are n + 1 terms in the expansion.

As long as n is relatively small, we can expand a given binomial directly by comparing it to the equivalent expansion of (a + b)n. For example:

Expand (x + 1)5.

and replacing a with x and b with 1 gives:

Using (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

(x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1

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Binomial expansions

Expand (2x – y)4.

and replacing a with 2x and b with –y gives:Using (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(2x – y)4 = (2x)4 + 4(2x)3(–y) + 6(2x)2(–y)2 + 4(2x)(–y)3 + (–y)4

= 16x4 – 32x3y + 24x2y2 – 8xy3 + y4

Notice that when the second term in a binomial is negative the signs of the terms in the expansion will alternate.

Suppose we wanted to expand (a + b)20.

We could find the 21st row of Pascal’s triangle, but this would take a very long time.

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Finding binomial coefficients

When n is large we can find the binomial coefficients using combinations theory.

Let’s look more closely at the expansion of

(a + b)4 = (a + b)(a + b)(a + b)(a + b)

aaaa

1 way

aaabaabaabaabaaa

4 ways 6 ways

abbbbabbbbabbbba

4 ways

bbbb

1 way

aabbabababbabbaabababaab

Ways of getting a4

Ways of getting a3b

Ways of getting a2b2

Ways of getting ab3

Ways of getting b4

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Finding binomial coefficients

The situation where no b’s (or four a’s) are chosen from any of the four brackets can be written as

The situation where one b (or three a’s) can be chosen from any of the four brackets can be written as:

4C0 or 4

0

.

The situation where two b’s (or two a’s) can be chosen from any of the four brackets can be written as

4C1 or 4

1

.

4C2 or . 4

2

This is the same as 4C4 or 4

4

.

This is the same as 4C3 or 4

3

.

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Finding binomial coefficients

The fifth row of Pascal’s triangle can be written as:

4

0

4

4

4

1

4

2

4

3

This corresponds to the values

1 4 6 4 1

The expansion of (a + b)4 can therefore be written as:

4 4 3 2 2 3 44 4 4 4 4( )

0 1 2 3 4a b a a b a b ab b

Or: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

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Finding binomial coefficients

The number of ways to choose r objects from a group of n objects is written as nCr and is given by

!

=! !

n n

r n rr

n! is read as ‘n factorial’ and is the product of all the natural numbers from 1 to n.

In general:

n! = n × (n –1) × (n – 2) × (n – 3) … × 2 × 1n! = n × (n –1) × (n – 2) × (n – 3) … × 2 × 1

n can also be 0 and by definition 0! = 1.

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Finding binomial coefficients

The value of n! gets large very quickly as the value of n increases. For example:

5! = 5 × 4 × 3 × 2 × 1 = 120

12! = 12 × 11 × 10 × … × 2 × 1 = 479 001 600

Fortunately, when we use the formula

to calculate binomial coefficients, many of the numbers cancel out. For example, for 4C2 we have

20! = 20 × 19 × 18 × … × 2 × 1 = 2 432 902 008 176 640 000

!

=! !

n n

r n rr

4 4!= =

2! 2!2

4×3×2×1=

(2×1)×(2×1)4×3

2×1= 6

2

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Finding binomial coefficients

Here are some more examples:

8 8!= =

3! 5!3

8×7×6×5× 4×3×2×1=

(3×2×1)×(5× 4×3×2×1)8×7×6

=3×2×1

56

This value corresponds to the number of ways of choosing 3 a’s from the 8 brackets in the expansion of (a + b)8.

56 is therefore the coefficient of a3b5 in the expansion of (a + b)8.

9 9!= =

7! 2!7

9×8×7×6×5× 4×3×2×1=

(7×6×5× 4×3×2×1)×(2×1)9×8

=2×1

36

This value corresponds to the number of ways of choosing 7 a’s from the 9 brackets in the expansion of (a + b)9.

36 is therefore the coefficient of a7b2 in the expansion of (a + b)9.

4

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Finding binomial coefficients

The effect of this cancelling gives an alternative form for nCr.

×( 1)×( 2)×...×( +1)=

!

n n n n n r

rr

In general, the expansion of (a + b)n can be written as:

A special case is the expansion of (1 + x)n

2 3(1+ ) = + + + +...+0 1 2 3

n nn n n nx x x x x

2 3( 1) ( 1)( 2)=1+ + + +...+

2! 3!nn n n n n

nx x x x

1 2 2( + ) = + + +...+0 1 2

n n n n nn n n na b a a b a b b

n

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Using the binomial theorem

Find the coefficient of a7b3 in the expansion of (a – 2b)10.

This method of finding the binomial coefficients is called the binomial theorem.

The term in a7b3 is of the form:

7 310( 2 ) =

3a b

7 310×9×8( 8 )

1×2×3a b4

= 120(–8a7b3)

So the coefficient of a7b3 in the expansion of (a – 2b)10 is –960.

= –960a7b3

3

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Using the binomial theorem

Use the binomial theorem to write down the first four terms in the expansion of (1 + x)7 in ascending powers of x.

7 2 37 7 7 7(1+ ) = + + + +...

0 1 2 3x x x x

2 37×6 7×6×5=1+ 7 + + +...

2×1 3×2×1x x x

3

How could we use this expansion to find an approximate value for 1.17?

=1+ 7 +x 221 +x 335 +...x

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Using the binomial theorem

To find an approximate value for 1.17 we can let x = 0.1 in the expansion

(1 + x)7 = 1 + 7x + 21x2 + 35x3 + …

This gives us

We can therefore leave out higher powers of x and still have a reasonable approximation.

1.17 ≈ 1 + 0.7 + 0.21 + 0.035

As 0.1 is raised to ever higher powers it becomes much smaller and so less significant.

1.17 ≈ 1 + 7 × 0.1 + 21 × 0.12 + 35 × 0.13

≈ 1.945

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Examination-style questions

Geometric sequences

Geometric series

The sum to infinity of a geometric series

Binomial expansions

Examination-style questions

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Examination-style question 2

a) Write down the first four terms in the expansion of (1 + ax)13 in ascending powers of x, where a > 0.

b) Given that in the expansion of (1 + ax)13 the coefficient of x is –b and the coefficient of x2 is 12b, find the value of a and b.

a) (1 + ax)13 =

= 1 + 13ax + 78a2x2 + 286a3x3 + …

2 313×12 13×12×111+13 + ( ) + ( ) +...

2 3×2ax ax ax

b) 13a = –b 1

78a2 = 12b 2

78a2 = 12 × –13a

Substituting into :1 2 78a = –156

a = –2

b = 26