Common Core Math 3 - 2020 Edition
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3.07 The binomial theorem
Lesson

In our investigation into Pascal's triangle we have seen it has connections to counting techniques and now it is time to formalize these connections. 

Our last activity was this one... 

Activity 8

Write down the following (on one line like I have)

4C0     4C1     4C2     4C3     4C4

Then underneath that write the answers.... 

What connection do you see to Pascal's triangle?

What was the connection? 

You should have discovered that the above answers to the combinations $\nCr{4}{r}$4Cr, formed an entire row in Pascal's triangle. 

In general, we can find the values of $\nCr{n}{r}$nCr in row number $n$n (starting at row $0$0) and the $r$r value is the element in the row, (also starting at $0$0).  

So the value for $\nCr{9}{4}$9C4, will be the row beginning  $1$1, $9$9, ..... and be the $5$5th number in the row - (remember we start the element from $0$0).

 

Worked example

Example 1

Find the missing elements in this row from Pascal's Triangle.

$1,9,$1,9, $\editable{A},84,\editable{B},\editable{C},84,36,9,1$A,84,B,C,84,36,9,1

Firstly we know that the lines of the triangle are symmetrical.  This helps us identify that box $\editable{A}$Ashould be the value of $36$36.  As reading from left to right is the same as reading from right to left. 

This symmetry doesn't help us with the values for $\editable{B}$B or $\editable{C}$C, but we can use our knowledge of combinations to solve this.  

$\editable{B}=\editable{C}$B=C because of of the symmetry.

$\editable{B}$B also equals the value of $\nCr{9}{4}$9C4 and $\editable{C}$C$=$=$\nCr{9}{5}$9C5, but we also know that $\nCr{9}{4}=\nCr{9}{5}$9C4=9C5 (confirming what we already knew from symmetry that the values will be the same).

$\editable{B}$B $=$= $\nCr{9}{4}$9C4 $=126$=126 

Thus both $\editable{B}$B and $\editable{C}=126$C=126.

 

Pascal's triangle and binomial distributions

 

$(a+b)^0$(a+b)0 $=$= $1$1
$(a+b)^1$(a+b)1 $=$= $a+b$a+b
$(a+b)^2$(a+b)2 $=$= $a^2+2ab+b^2$a2+2ab+b2
$(a+b)^3$(a+b)3 $=$= $a^3+3a^2b+3ab^2+b^3$a3+3a2b+3ab2+b3
$(a+b)^4$(a+b)4 $=$= $a^4+4a^3b+6a^2b^2+4ab^3+b^4$a4+4a3b+6a2b2+4ab3+b4
$(a+b)^5$(a+b)5 $=$= $a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$a5+5a4b+10a3b2+10a2b3+5ab4+b5

 

Consider the distributions above of $(a+b)^n$(a+b)n.  Particularly note the following patterns.

  • For each distribution to the power $n$n, there are $n+1$n+1 elements.
  • For each term, the sum of the exponents is $n$n.  
  • Powers of $a$a decrease from left to right, from $n$n down to $0$0.
  • Powers of $b$b increase from left to right, from $0$0 up to $n$n
  • The coefficients start at $1$1, end at $1$1, and are the terms of the relevant row from Pascals triangle.

 

Worked example

Example 2

What are the coefficients for the distribution of $(a+1)^7$(a+1)7, and then write out the full distribution. 

So we can see that we will have $n+1=8$n+1=8 terms.

We can refer to the relevant row in Pascal's triangle, specifically this row 

This shows us that the coefficients will be 

$1,7,21,35,35,21,7,1$1,7,21,35,35,21,7,1.

Thus the full distribution of $(a+1)^7$(a+1)7 will be

  $\left(a+1\right)^7$(a+1)7
$=$= $a^7+7a^61^1+21a^51^2+35a^41^3+35a^31^4+21a^21^5+7a^11^6+1^7$a7+7a611+21a512+35a413+35a314+21a215+7a116+17
$=$= $a^7+7a^6+21a^5+35a^4+35a^3+21a^2+7a+1$a7+7a6+21a5+35a4+35a3+21a2+7a+1

 

The binomial theorem

We can concisely summarize the pattern in distributions we have observed as a formula called the binomial theorem. This formula will also allow us to find particular terms in an distribution.

Using our knowledge that for an distribution 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)

This results in the distribution 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.  

Worked examples

Example 3

Distribute $(2x+3)^5$(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 4

What is the seventh term in the distribution 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.

 

Practice questions

QUESTION 1

You are given some of the entries in a particular row of Pascal’s triangle. Fill in the missing entries.

  1. $1$1 , $8$8 , $\editable{}$ , $56$56 , $70$70 , $\editable{}$ , $28$28 , $\editable{}$ , $1$1

QUESTION 2

How many terms are there in the distribution of $\left(m+y\right)^8$(m+y)8?

QUESTION 3

Using the relevant row of Pascal’s triangle, determine the coefficient of each term in the distribution of $\left(5+b\right)^5$(5+b)5.

  1. $\left(5+b\right)^5$(5+b)5$=$=$\editable{}$$\times$×$5^5b^0$55b0$+$+$\editable{}$$\times$×$5^4b^1$54b1$+$+$\editable{}$$\times$×$5^3b^2$53b2$+$+$\editable{}$$\times$×$5^2b^3$52b3$+$+$\editable{}$$\times$×$5^1b^4$51b4$+$+$\editable{}$$\times$×$5^0b^5$50b5.

QUESTION 4

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

  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 5

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

Outcomes

A.APR.C.5

(+) Know and apply the binomial theorem for the expansion of (x y)^n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by pascal's triangle

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