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Binomial Expansion

Binomial Expansion

A binomial expansion is a method used to allow us to expand and simplify algebraic expressions in the form \( (x+y)^n\) into a sum of terms of the form \(ax^by^c\). If \(n\) is an integer, \(b\) and \(c\) also will be integers, and \(b + c = n\).

We can expand expressions in the form \( (x+y)^n\) by multiplying out every single bracket, but this might be very long and tedious for high values of \(n\) such as in \( (x+y)^{20}\) for example. This is where using the Binomial Theorem comes in useful.

The binomial theorem

The binomial theorem allows us to expand an expression of the form \( (x+y)^n\) into a sum. A general formula for a binomial expression is:

\[ (x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n.\]

Which can be simplified to:

\[ \begin{align} (x+y)^n &= \sum\limits_{k=0}^n \binom{n}{k} x^{n-k}y^k \\ &= \sum\limits_{k=0}^n \binom{n}{k} x^ky^{n-k} . \end{align}\]

Where both \(n\) and \(k\) are integers. This is also known as the binomial formula. The notation

\[ \binom{n}{k}\]

can be referred to as '\(n\) choose \(k\)' and gives a number called the binomial coefficient which is the number of different combinations of ordering \(k\) objects out of a total of \(n\) objects. The equation for the binomial coefficient (\(n\) choose \(k\) or \(^nC_r\) on a calculator) is given by:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!}\]

Where '!' means factorial. Factorial means the product of an integer with all the integers below it. For example for \(5\) choose \(3\), we would have:

\[ \begin{align} \binom{5}{3} &= \frac{5!}{3!(5-3)!} \\ &= \frac{5\cdot 4\cdot 3 \cdot 2 \cdot 1}{(3\cdot 2\cdot 1)(2\cdot 1)} \\ &= 10. \end{align}\]

How do you do a binomial expansion?

To understand how to do a binomial expansion, we will look at an example. Let's say we want to expand \( (x+y)^4\). In this case, \(n = 4\) and \(k\) will vary between \(0\) and \(4\). Using the formula for the binomial expansion, we can write:

\[ (x+y)^4 = \binom{4}{0}x^4y^0 + \binom{4}{1}x^3y^1 + \binom{4}{2}x^2y^2 + \binom{4}{3}x^1y^3+\binom{4}{4}x^0y^4.\]

We can now use the equation for the binomial coefficient to find all the constant terms in this expression. For the first term we have:

\[ \begin{align} \binom{4}{0} &= \frac{4!}{0!(4-0)!} \\ &= \frac{4 \cdot 3\cdot 2\cdot 1}{1\cdot (4 \cdot 3\cdot 2\cdot 1 )} \\ &= 1. \end{align} \]

Repeating this for all five coefficients, we end up with binomial coefficients of \(1\), \(4\), \(6\), \(4\), \(1\) in order. Therefore, our expression for the binomial expansion simplifies to:

\[ x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4.\]

Note that \(y\) could also be replaced by any number.

Binomial expansion for fractional and negative powers

Sometimes you will encounter algebraic expressions where n is not a positive integer but a negative integer or a fraction. Let's consider the expression \(\sqrt{1-2x}\) which can also be written as

\[ (1- 2x)^\dfrac{1}{2} \] where \(x < 0.5\). In this case, it becomes hard to find the formula to find the binomial coefficients,

because we can't find the factorials for a negative or rational number. However, if we look at an example for a positive integer, we can find a more general expression that we can then also apply to negative and fractional numbers. For example for

\[ \binom{6}{3}\]

we have

\[ \begin{align} \binom{6}{3}&= \frac{6!}{3!(6-3)!} \\ &= \frac{6\cdot 5\cdot 4}{3!} \\ &= \frac{6(6-1)(6-2)}{3!}. \end{align}\]

From this we observe that

\[ \binom{n}{k} = \frac{n(n-1)(n-2)(n-3)\dots (n-k+1)}{k!} \]

and therefore the more general expression for the binomial theorem is the infinite formula

\[ (a+b)^n = \frac{a^n}{0!} + \frac{na^{n-1}b}{1!} + \frac{n(n-1)a^{n-2}b^2}{2!} + \frac{n(n-1)(n-2)a^{n-3}b^3}{3!} + \dots \]

Let's look at \(\sqrt{1-2x}\). In this case \(a = -2x\), \(b = 1\) and \(n =1/2\). Substituting this we get:

\[ \begin{align} \frac{(-2x)^\frac{1}{2}}{0!} &+ \frac{\left(-\frac{1}{2}\right) (-2x)^{-\frac{1}{2}}\cdot 1 }{1!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) (-2x)^{-\frac{3}{2}}\cdot 1^2 }{2!} \\ &\quad + \frac{\left(-\frac{1}{2}\right) \left(-\frac{1}{2}\right) \left(-\frac{3}{2}\right) (-2x)^{-\frac{5}{2}}\cdot 1^3 }{3!} + \dots \end{align}\]

Using Mac Laurin's expansion we can say that the above expression converges to

\[ \sqrt{1-2x} = 1 - x - \frac{x^2}{2} - \frac{x^3}{2}.\]

Binomial Expansion - Key takeaways

  • A binomial expansion helps us to simplify algebraic expressions into a sum
  • The formula for the binomial expansion is:

    \[ (x+y)^n = \binom{n}{0}x^ny^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n-1}x^1y^{n-1} + \binom{n}{n}x^0y^n.\]

  • The binomial coefficients or constant terms in this expression are found using:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!}\]
  • \[ (1+a)^n = 1 + na+ \frac{n(n-1)}{2!}a^2 + \frac{n(n-1)(n-2)}{3!}a^3 + \dots \]

Frequently Asked Questions about Binomial Expansion

The constant term is found by using the formula

 n choose k=n!/k!(n-k)!

A binomial expansion is a method that allows us to simplify complex algebraic expressions into a sum.

You can use the binomial expansion formula

(x+y)^n=(nC0)x^n y^0+(nC1)x^/n-1)y^1+(nC2)x^(n-2)y^2+...+(nCn-1)x^1y^(n-1)+(nCn)x^0y^n

Final Binomial Expansion Quiz

Question

What is the binomial expansion used for?

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Answer

Expanding out things like \((x+y)^n\) without having to do all the multiplication.

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Question

 What is the simplified form of the binomial expansion formula using summation notation?

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Answer

\[(x+y)^n = \sum\limits_{k=0}^n {n\choose k} x^{n-k}y^k.\]


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Question

How do you find the binomial coefficients?

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Answer

Using \( n\choose k\).

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Question

What is 6 choose 3? 


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Answer

20.

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Question

What is 7 choose 4?

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Answer

35.

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Question

What is 5 choose 0?

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Answer

1.

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Question

What is 6 choose 6? 


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Answer

1.

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Question

Expand \((1+2x)^5\).


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Answer

\(1+10x+40x^2+80x^3+80x^4+32x^5\).


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Question

Which of these is the correct expansion of \((x^2-y)^2\)?

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Answer

\(x^4-2x^2y+y^2\).


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Question

Find the first three terms in the expansion of \((4+5x)^{\frac{1}{2}}\).

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Answer

\(2+\dfrac{5}{4}x -\dfrac{25}{64}x^2\).


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Question

Find the first three terms in the expansion of \((1-2x)^{-\frac{1}{2}}\).

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Answer

\(1+x+\dfrac{3}{2}x^2\).


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Question


Given that the coefficient of \(x^2\) in the expansion of \((1+ax)^7\) is \(525\), find the possible values of \(a\).




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Answer

\(5\).

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Question

Which of these is equivalent to \(n!\)?

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Answer

\(n (n-1)\dots (2)(1) \).

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Question

Which of these is the correct expansion of \(6 choose 2\)?

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Answer

\( \dfrac{6!}{2!4!}\).

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Question

How do you read the notation \(n \choose k\)?

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Answer

\(n\) choose \(k\).

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Question

What does the notation \(n\choose k\) mean?

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Answer

\(\dfrac{n!}{(n-k)!k!}\).

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Question

Functions in mathematics with the symbol (!) that multiply a number by every number that precedes it are…?

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Answer

Factorials

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Question

What is the factorial notation?

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Answer

n!. Where n is a positive integer.

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Question

The rule says that the factorial of any number is that number times the factorial of (that number minus 1). n! = n × (n−1) is called…?


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Answer

Factorial rule

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Question

What is 0!?


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Answer

0! = 1

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Question

What are factorials used for?

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Answer

Arrangements, permutations, and combinations.


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Question

What is the factorial of 9?


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Answer

9! =9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362,880

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Question

How many possible combinations can you make with a four-digit number?


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Answer

4 × 3 × 2 × 1 = 26

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Question

How do you calculate factorials?


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Answer

Multiply your number by every number below it.

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Question

Evaluate 3!7!

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Answer


 3!7! = (3× 2× 1) (7 × 6 × 5 × 4 × 3 × 2 × 1) = 30240

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Question

What can Pascal´s triangle help us find?

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Answer

Binomial coefficients.

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Question

What is the \(n\) number for the first row in Pascal´s triangle?

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Answer

\(n=0\).

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Question

What is the first value in Pascal's triangle?

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Answer

\(1\).

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Question

What are the values on the extremes of each row?

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Answer

\(1\).

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Question

How do you find the values of a row?

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Answer

By adding together the values above it.

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Question

How many elements are in row \(n\) of Pascal's triangle?

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Answer

\(n+1\).

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Question

What is the formula for finding the sum of each row?

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Answer

\(2^n\).

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Question

How do you find Fibonacci´s sequence in Pascal's triangle?

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Answer

By adding the values diagonally.

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Question

What are the first five values of Fibonacci's sequence?

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Answer

1, 1, 2, 3, 5.

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Question

What is the sum of the elements in the 8th row (\(n=7\)) of Pascal's triangle?

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Answer

\(128\).

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Question

What are the binomial coefficients for the expansion \((x+y)^6\)? 

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Answer

\(1, 6, 15, 20, 15, 6, 1\).

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Question

What are the binomial coefficients for the expansion \( (x+2y)^4\).

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Answer

\(1, 4, 6, 4, 1\).

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Question

How do you use Pascal's Triangle to find the Fibonacci numbers?

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Answer

Sum along the diagonals.

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Question

If you are making Pascal's triangle from scratch, what three numbers do you put in a triangle first?

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Answer

\(1\) in all three spots.

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Question

If you want to find the next row of Pascal's triangle, what do you do to the elements of the previous row?

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Answer

Add them.

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