© Boardworks Ltd 2005 1 of 39 Contents © Boardworks Ltd 2005 1 of 39 Binomial expansions Geometric...
-
Upload
harrison-woodbridge -
Category
Documents
-
view
226 -
download
1
Transcript of © Boardworks Ltd 2005 1 of 39 Contents © Boardworks Ltd 2005 1 of 39 Binomial expansions Geometric...
© Boardworks Ltd 20051 of 39
Co
nte
nts
© Boardworks Ltd 20051 of 39
Binomial expansions
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
© Boardworks Ltd 20052 of 39
Pascal’s Triangle
© Boardworks Ltd 20053 of 39
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?
© Boardworks Ltd 20054 of 39
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
© Boardworks Ltd 20055 of 39
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.
© Boardworks Ltd 20056 of 39
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
© Boardworks Ltd 20057 of 39
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
.
© Boardworks Ltd 20058 of 39
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
© Boardworks Ltd 20059 of 39
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.
© Boardworks Ltd 200510 of 39
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
© Boardworks Ltd 200511 of 39
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
© Boardworks Ltd 200512 of 39
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
© Boardworks Ltd 200513 of 39
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
© Boardworks Ltd 200514 of 39
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
© Boardworks Ltd 200515 of 39
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
© Boardworks Ltd 200516 of 39
Co
nte
nts
© Boardworks Ltd 200516 of 39
Examination-style questions
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
© Boardworks Ltd 200517 of 39
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