New Zealand
Level 8 - NCEA Level 3

# Binomial Expansions and Coefficients

Lesson

Our final step in this journey is to connect Pascal's Triangle. How are the rows of the triangle connected to the combinations $\nCr{n}{r}$nCr? How can the rows identify the value of the coefficients in a binomial expansion?

What we get when we combine them is the knowledge that for an expansion of $\left(a+b\right)^n$(a+b)n the coefficients will be dictated by the combinations of $\nCr{n}{0}$nC0, $\nCr{n}{1}$nC1, $\nCr{n}{2}$nC2, $\dots$, $\nCr{n}{n}$nCn, also notated as $\binom{n}{0}$(n0),$\binom{n}{1}$(n1),$\binom{n}{2}$(n2),$...$...,$\binom{n}{n}$(nn)

So this results in the expansion looking like this

$(a+b)^n=$(a+b)n=$\binom{n}{0}$(n0)$a^n$an$+$+$\binom{n}{1}$(n1)$a^{n-1}b^1+$an1b1+$\binom{n}{2}$(n2)$a^{n-2}b^2+$an2b2+$\binom{n}{3}$(n3)$a^{n-3}b^3+...+$an3b3+...+$\binom{n}{r}$(nr)$a^{n-r}b^r+...+$anrbr+...+$\binom{n}{n-1}$(nn1)$a^1b^{n-1}+$a1bn1+$\binom{n}{n}$(nn)$b^n$bn

Thus any particular term can be found using $\binom{n}{r}$(nr)$a^{\left(n-r\right)}$a(nr)$b^r$br.

##### Example 1

Expand $(2x+3)^5$(2x+3)5.

$(a+b)^n=$(a+b)n=$\binom{n}{0}$(n0)$a^n+$an+$\binom{n}{1}$(n1)$a^{n-1}b^1+$an1b1+$\binom{n}{2}$(n2)$a^{n-2}b^2+$an2b2+$\binom{n}{3}$(n3)$a^{n-3}b^3+...+$an3b3+...+$\binom{n}{r}$(nr)$a^{n-r}b^r+...$anrbr+...$\binom{n}{n-1}$(nn1)$a^1b^{n-1}+$a1bn1+$\binom{n}{n}$(nn)$b^n$bn

$(2x+3)^5=$(2x+3)5=$\binom{5}{0}$(50)$(2x)^5+$(2x)5+$\binom{5}{1}$(51)$(2x)^{5-1}3^1+$(2x)5131+$\binom{5}{2}$(52)$(2x)^{5-2}3^2+$(2x)5232+$\binom{5}{3}$(53)$(2x)^{5-3}3^3+$(2x)5333+$\binom{5}{4}$(54)$(2x)^13^{5-1}+$(2x)1351+$\binom{5}{5}$(55)$(3)^5$(3)5

$(2x+3)^5=1(2x)^5+5(2x)^43^1+10(2x)^33^2+10(2x)^23^3+5(2x)^13^4+1(3)^5$(2x+3)5=1(2x)5+5(2x)431+10(2x)332+10(2x)233+5(2x)134+1(3)5

$(2x+3)^5=32x^5+15(16x^4)+90(8x^3)+270(4x^2)+810x+243$(2x+3)5=32x5+15(16x4)+90(8x3)+270(4x2)+810x+243

$(2x+3)^5=32x^5+240x^4+720x^3+1080x^2+810x+243$(2x+3)5=32x5+240x4+720x3+1080x2+810x+243

##### Example 2

What is the seventh term in the expansion of $(m-2n)^{12}$(m2n)12?

We need to construct the seventh term from this $\binom{n}{r}$(nr)$a^{\left(n-r\right)}$a(nr)$b^r$br where $n$n is $12$12 and $r$r is $6$6

The coefficient $\binom{n}{r}$(nr) where $n$n is $12$12 and $r$r is $6$6 is $\binom{12}{6}=924$(126)=924.

The term will have both $m$m and $(2n)$(2n) components. The $m$m component would be $m^{12-6}=m^6$m126=m6

The $2n$2n component would be $(2n)^6=64n^6$(2n)6=64n6.

So putting that altogether will give us $924m^6\times64n^6=59136m^6n^6$924m6×64n6=59136m6n6.

#### More Worked Examples

##### QUESTION 1

How many terms does the expansion of $\left(9x+6y\right)^8$(9x+6y)8 have?

##### QUESTION 2

Using the binomial theorem, determine the missing powers in the following expansion.

1. $\left(4p+3q\right)^3=\nCr{3}{0}\left(4p\right)^{\editable{}}\left(3q\right)^0+\nCr{3}{1}\left(4p\right)^{\editable{}}\left(3q\right)^1+\nCr{3}{2}\left(4p\right)^{\editable{}}\left(3q\right)^2+\nCr{3}{3}\left(4p\right)^{\editable{}}\left(3q\right)^3$(4p+3q)3=3C0(4p)(3q)0+3C1(4p)(3q)1+3C2(4p)(3q)2+3C3(4p)(3q)3

##### QUESTION 3

Expand $\left(\sqrt{2}x+\frac{1}{y}\right)^4$(2x+1y)4.

### Outcomes

#### M8-3

Use permutations and combinations