7.06 Pascal's triangle

In our investigation into Pascal's triangle we found that we can generate the numbers in the rows of the triangle using our calculator.

 

What is the connection? This notation is used for what is called combinations in mathematics. 

You should have discovered that the above answers to the combinations $\binom{4}{3}$(43), form an entire row in Pascal's triangle. 

In general, we can find the values of $\binom{4}{3}$(43) 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 $\binom{9}{4}$(94), will be the row beginning  $1$1, $9$9, ..... and be the $5$5^th number in the row - (remember we start the element from $0$0).

 

Worked example

Example 1

Find the missing elements in the 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 $\binom{9}{4}$(94) and $\editable{C}$C$=$=$\binom{9}{5}$(95), but we also know that $\binom{9}{4}=\binom{9}{5}$(94)=(95) (confirming what we already knew from symmetry that the values will be the same).

$\editable{B}$B $=$= $\binom{9}{4}$(94) $=126$=126 

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

 

Pascal's triangle and binomial expansions

 

$(a+b)^0$(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 expansions above of $(a+b)^n$(a+b)n.  Particularly note the following patterns.

• For each expansion 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

Determine the coefficients for the expansion of $(a+1)^7$(a+1)7, and then write out the full expansion. 

So we can see that we will have $n+1=8$n+1=8 terms. As it is $+1$+1 in the bracket and $1^n=1$1n=1, the numbers from Pascal's triangle will be the coefficients. 

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 expansion 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

 

Reflect: These coefficients could also have been generated by evaluating $\binom{7}{0}$(70), $\binom{7}{1}$(71), $\binom{7}{2}$(72), $\binom{7}{3}$(73), $\binom{7}{4}$(74), $\binom{7}{5}$(75), $\binom{7}{6}$(76), $\binom{7}{7}$(77)

The binomial theorem

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

Using our 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)

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)$a^{n-1}b^1+$an−1b1+$\binom{n}{2}$(n2)$a^{n-2}b^2+$an−2b2+$\binom{n}{3}$(n3)$a^{n-3}b^3+...+$an−3b3+...+$\binom{n}{r}$(nr)$a^{n-r}b^r+...+$an−rbr+...+$\binom{n}{n-1}$(nn−1)$a^1b^{n-1}+$a1bn−1+$\binom{n}{n}$(nn)$b^n$bn

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

Worked examples

Example 3

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

$(a+b)^n=$(a+b)n=$\binom{n}{0}$(n0)$a^n+$an+$\binom{n}{1}$(n1)$a^{n-1}b^1+$an−1b1+$\binom{n}{2}$(n2)$a^{n-2}b^2+$an−2b2+$\binom{n}{3}$(n3)$a^{n-3}b^3+...+$an−3b3+...+$\binom{n}{r}$(nr)$a^{n-r}b^r+...$an−rbr+...$\binom{n}{n-1}$(nn−1)$a^1b^{n-1}+$a1bn−1+$\binom{n}{n}$(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)5−131+$\binom{5}{2}$(52)$(2x)^{5-2}3^2+$(2x)5−232+$\binom{5}{3}$(53)$(2x)^{5-3}3^3+$(2x)5−333+$\binom{5}{4}$(54)$(2x)^13^{5-1}+$(2x)135−1+$\binom{5}{5}$(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 expansion of $(m-2n)^{12}$(m−2n)12?

We need to construct the seventh term from this $\binom{n}{r}$(nr)$a^{\left(n-r\right)}$a(n−r)$b^r$br where $n$n is $12$12 and $r$r is $6$6 (remember that $\binom{12}{0}$(120) is the coefficient of the first term). 

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 $m^{12-6}=m^6$m12−6=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

QUESTION 2

QUESTION 3