The Binomial Expansion

• Edexcel, Module - C2, Chapter- Sequences and Series
• AQA, Module - C2, Chapter - Sequences and Series
• OCR, Module - C2, Chapter - Sequences and Series

 The Sequence and Series chapter in c2, is quite big, so I will divide it into 3 / 4 posts. a) Binomial Expansion 1 b) Binomial Expansion 2, c) Geometric Sequences 1 ,d) Geometric Sequences 2. 

The Binomial Expansion, is a theorem which allows us to expand (a + b)^n, where n is an integer. (a and b are just what i've used it, they can be any letters).

Why we need it ?

Let's say we have to expand the following terms :

(x + y)^1 = x + y
(x + y)^2 = x^2 + 2xy + y^2
(x + y)^3 = (x + y)(x + y)^2
               = (x + y)(x^2 + 2xy + y^2)
               = x^3 + 3x^2y + 3xy^2 + y^3

As we see, when increasing the power, (n) we get more and more terms, and it becomes more and more confusing, say we had to expand (x + y)^7, this would mean we would get 8 terms, after expanding the bracket 7 times, and then simplifying it. This would be a tedious and extremely long !

The Binomial theorem therefore allows us to expand the binomials ( two terms e.g. x + y or x + 3y or a - 3b). In each term of the expansion, there will be a coefficient of the term, these coefficients are either determined by Pascal's Triangle or Factorial Formula.

Let's see how this works, with the examples above... Let's expand (x + y)^3 using the theorem :

Binomial Theorem = To expand the binomial (x + y)^n, it can be written in the form :

This is the binomial coefficient, where n is the power and k is the increasing value till we reach n.

Let's use take our example using the theorem :

(x + y)^2 = (2,0) x^2 y^0 + (2,1) x^1 y^1 + (2,2) x^0 y^2

*The brackets are the binomial coefficients in blue , which are calculated using the formula :

* Notice the x terms start with the power n (in this case 2), and decrease by one power the next term, till the reach the power 0, simultaneously the y terms start with the power 0, and increase by one power the next term, till they reach the power n.

* We finish the binomial expansion when the binomial coefficient is (n,n) in this case (2,2), and when the x term has reached to the power 0 (which is 1), and the y term is to power n (in this case 2) . (Remember anything to the power 0 = 1 and anything to the power 1 = itself )


Let's simplify it down further, these are the binomial coefficients for each term :

! = Factorial (it calculates the product of all the positive integers less than or equal to some integer n)
(e.g. 4! = 4 x 3 x 2 x 1 = 24)

Note :  1! = 1 and  0! = 1

(2,0) = 2! / [ 0! (2 -0)! ]
         = 2 / [ 1 x 2 ]
         = 2 / 2
         = 1

(2,1) = 2! / [ 1! (2 -1)! ]

         = 2 / [ 1 x 1 ]
         = 2 /1
         = 2

(2,2) = 2! / [ 2! (2 -2)! ]

         = 2 / [ 2 x 1]
         = 2 / 2
         =1

So we have calculated the binomial coefficients for the terms, giving us :

(x + y)^2 = x^2 + 2xy + y^2


Pascal's Triangle ( Second way to find binomial coefficients - helps when n is small)

Another way to find the binomial cofficients, is the Pascal's Triangle. We start with n = 0, and start with number 1. Each row, will list the binomial cofficients in order of the expansion, each starting with 1 and ending with 1, depending on n.

To find the coefficient underneath, two gaps, we simply add the coefficients above it.

n = 0                                                                              1

n = 1                                                                  1                     1

n = 2                                                      1                       2                   1

n = 3                                         1                       3                      3                     1

n = 4                            1                        4                       6                   4                         1

n = 5               1                        5                      10                    10                    5                          1

* The Triangle goes on as n increases, and the number of coefficients is always ( n + 1)
* The Triangle is symmetrical halfway, so we see that the first half of the coefficients of terms, is always the same as the second half
* Each row of the triangle starts and ends with 1 (meaning the first coefficient and last coefficient of any expansion is always 1)

Double check the expansion (x + y)^2 , where n = 2,

Check the row n = 2

The binomial coefficients should be 1, 2 and 1

1x^2 + 2xy + 1y^2

So we've seen how to expand expressions in the form (x + y)^n, and use two methods of working out the binomial coefficients. In the next post I will show you a couple of examples, and another expansion theorem, which allows us to expand algeabraic expressions in the form (1 + x)^ n.

Sequences and Series

•  Edexcel - C1, Sequences and Series
• AQA - C2, Sequences and Series
• OCR - C2, Sequenes and Series

A sequence is a list of numbers, in it's basic definition. This list can be of a finite / infinite length.e.g. 2,4,6,8.. is a sequence of all positive even numbers. There are many types of sequences, we shall be looking at the Arithmetic Series / Progression.

A series is the sum of the terms in a sequence. Again these can be finite and infinite, depending on the sequence itself.

Arithmetic Sequences

An Arithmetic Sequence is a sequence of numbers, such that difference between the terms is a constant. e.g. 5,9,13,17,21 .. this difference here is + 4. This difference is called the common difference.


Each number in the sequence, is called a term. We call the first term (u1), second term (u2), third term (u3)... and so on. This is just notation. In the previous sequence example, 5 would be the first term, 9 the second, 13 the third.. and so on. The first term of a sequence is a. While the common difference is d.

Nth term, is a rule for finding any term in the sequence. Say if i wanted to find the 28th term of the previous sequence, it would be very long to add 4 each time, to get till the 28th term. Instead we can form a rule to find any term in that sequence. The formula for finding the nth term of a sequence is :

U(n) = a + (n-1)d

a = first term
d= common difference
n = the term you're finding

*This formula will work for any Arithmetic Sequence.

How we got this formula ?
As you know the first term is a. If each term goes up by a common difference, the second term must be a + d. The third term must be a + d + d = a + 2d, The fourth term must be a + d + d + d = a +3d.... and so on... if we look for the nth term it must be a + (n-1)d.


Example
a) Say we have an arithmetic sequence with the first term being 9. The common difference is -4. Find the 80th term ?

Here a is 9, d = -4 , and n=80

Use the formula U(80) = 9 + (80-1)*-4
=9 + (79*-4)
= 9+-(316)
= -307

Arithmetic Series

We also need to be able to find the Sum of an Arithmetic Sequence. There is a formula, we also need to be able to prove that formula (abit confusing, i'll include it in my image notes).

S(n) = The sum of n terms
n = the number of terms
a1 = first term (a)
d = common difference

Example

2,5,8,11,14,17....
a) Find the sum of the first ten terms
b) Find the sum of the terms starting from the 11th term and ending with the 28th term.

Identify a and d.
a = 2, d = +3

a) Use the Formula : 
S(10) =  10/2 [ 2(2) + (10-1)3]
= 5 [4 + 9(3)]
= 5 [4 + 27]
= 5 *31
=155

b) Use the Formula, note a is different.

They want to start from the 11th term, so the 11th term will be a.
 Use the nth term formula to find the 11th term : (Here we use a as 2)

=2 + (11-1)*3
= 2 + (10)*3
= 2 + 30
=32

11th term is 32, which is the a

Count how many terms are from 11 to 28. (18 terms), Now use the formula :
S(18) = 18/2 [ 2(32) + (18-1)*3]
= 9 [64 + 17*3]
= 9 [64 + 51]
= 9 x115
= 1035

Sigma Notation

The sigma sign Σ, is another notation you need to be able to interpret. It is simply the summation of an arithmetic sequence. I will use an example to make you understand it:

Here is an example, for sigma notation. The actual sign Σ means the sum. Now k^2, is the function / rule. The first term we have to add is below the sign, which is 2. So a =2, we have to add all the terms from 2 till 6. The number above the sign, is where we end. So the sum is 2 squared + 3 squared + 4 squared + 5 squared + 6 squared.. = 90.

Proof for the Sum of Arithmetic Series

C1 Introduction

C1. The first module of AS, of any Alevel Maths Course. Suppose to be the easiest, it generally forms on from the A-A* topics of gcse. However, you don't have a calculator... so brush up on general arithmetic skills, and fractions. Topics :

1)  Algebra and Functions
*Laws of Indices, Surds, Rationalising the Denominator, Quadratic Equations (Discriminant, Completing the Square, Quadratic Formula,Factorisation), Sketching Graphs and Transformations of Graphs.

2) Coordinate Geometry
*Equation of a Straight Line, Gradients, Perpendicular Lines

3) Sequences and Series
*Terms, nth Term, Arithmetic Series, Sum of Terms and Stigma Notations.

4)Differentiation
*Differentiating basic functions, and relation to Gradients.

5)Integration
*Indefinite Integration