9.07 Binomial expansion
What is Pascal's triangle?
Pascals triangle is a triangular array (which just means triangular shaped group of numbers) that have many special patterns and properties.
Pascal's triangle is named after a French mathematician called Blaise Pascal, although mathematicians from around the world for centuries before him had been using the patterns contained within the triangle. The triangle is constructed using a summation process.
We start by setting out our triangular array, the template at the end of this page might help you if you want to print it out.
Put 1's all the way down the left and right sides.
The third row, has a blank gap. We get the value for this by adding the two directly above it.
which is
The next row continues in the same manner, adding the two numbers directly above it...
which becomes 
We can continue on and on in this pattern. This creates our Pascal's triangle.
Worked example
example 1
a) Complete the following table:
| $\nCr{n}{r}$nCr | $\nCr{4}{0}$4C0 | $\nCr{4}{1}$4C1 | $\nCr{4}{2}$4C2 | $\nCr{4}{3}$4C3 | $\nCr{4}{4}$4C4 |
|---|---|---|---|---|---|
| Value |
b) What do you notice when you compare the values to Pascal's triangle?
a)
Think: You can use your calculator to find the values.
Do:
| $\nCr{n}{r}$nCr | $\nCr{4}{0}$4C0 | $\nCr{4}{1}$4C1 | $\nCr{4}{2}$4C2 | $\nCr{4}{3}$4C3 | $\nCr{4}{4}$4C4 |
|---|---|---|---|---|---|
| Value | $1$1 | $4$4 | $6$6 | $4$4 | $1$1 |
b) When comparing the values in the table to Pascal's triangle, we notice that the values are the same as the 5th row in Pascal's triangle:
Pascal's triangle has connections to counting techniques.
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) using Pascal's triangle:
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 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 $\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 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
What are 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.
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 |
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.
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
You are given some of the entries in a particular row of Pascal’s triangle. Fill in the missing entries.
$1$1 , $8$8 , $\editable{}$ , $56$56 , $70$70 , $\editable{}$ , $28$28 , $\editable{}$ , $1$1
QUESTION 2
How many terms are there in the expansion 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 expansion of $\left(5+b\right)^5$(5+b)5.
$\left(5+b\right)^5$(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 expansion.
$\left(4p+3q\right)^3=\nCr{3}{0}\left(4p\right)^{\editable{}}\left(3q\right)^0+\nCr{3}{1}\left(4p\right)^{\editable{}}\left(3q\right)^1+\nCr{3}{2}\left(4p\right)^{\editable{}}\left(3q\right)^2+\nCr{3}{3}\left(4p\right)^{\editable{}}\left(3q\right)^3$(4p+3q)3=3C0(4p)(3q)0+3C1(4p)(3q)1+3C2(4p)(3q)2+3C3(4p)(3q)3
QUESTION 5
Expand $\left(\sqrt{2}x+\frac{1}{y}\right)^4$(√2x+1y)4.