Binomial Expansion Part1

From StpmWiki

Notes

Compared to many other subtopics, binomial expansion can be considered a good prospect to score in STPM, as questions don't usually run too far from the standard questions usually asked. That said, there are many pitfalls if we are not careful. Good understanding of the methods are essential, as well as developing systematic work flows to minimize careless mistakes.

Learning Objectives (Syllabus)

expand ${\left(a+b \right)}^{n}$ where $n \,$ is a positive integer
expand ${\left(1+x \right)}^{n}$ where $n \,$ is a rational number and $\left|x \right|<1$
use the binomial expansions for approximation

Prior Knowledge

Strong basics in index is a MUST in this topic
Sequence
Surds, factorials, inequalities are required in a few places
As with all subtopics in chapter 3, ability to recognize pattern is essential

Binomial Coefficient

$\binom{n}{r}$ $=\frac{n!}{\left(n-r\right)!r!}$ where $n,r \,$ are positive integers and $0\leq r\leq n$

Note that this is the same definition as ${}^{n}{C}_{r} \,$ used in probability. In fact we will use the ${}^{n}{C}_{r} \,$ function in the calculator to find the value of $\binom{n}{r}$ .

Examples:

$\binom{10}{3}$ $=\frac{10!}{7!3!}$
To find it's value, however, we just need to use calculator and don't need to write out the working $\binom{10}{3} = 120$

Important values

$\binom{n}{0}$ $=\frac{n!}{\left(n-0\right)!0!}$ $=\frac{n!}{n!1}=1$
$\binom{n}{1}$ $=\frac{n!}{\left(n-1\right)!1!}$ $=\frac{n\left(n-1\right)!}{\left(n-1\right)!}=n$
$\binom{n}{n-1}$ $=\frac{n!}{\left[n-\left(n-1\right)\right]!\left(n-1\right)!}$ $=\frac{n\left(n-1\right)!}{1!\left(n-1\right)!}=n$
$\binom{n}{n}$ $=\frac{n!}{\left(n-n\right)!n!}$ $=\frac{n!}{0!n!}=1$

In other words

$\binom{n}{0}=\binom{n}{n}=1$ For example, $\binom{5}{0}=1, \binom{5}{5}=1$
$\binom{n}{1}=\binom{n}{n-1}=n$

Expansion for positive integer n

${\left(a+b \right)}^{n}$ $={a}^{n}+\binom{n}{1}{a}^{n-1}b+\binom{n}{2}{a}^{n-2}{b}^{2}+\dots+\binom{n}{n-1}a{b}^{n-1}+{b}^{n}$

Got it? Let's see it step by step

- Start with ${a}^{n} \,$
- The next term,
- After that,
- Try writing out the next term $\binom{n}{3}{a}^{n-3}{b}^{3}$
- And the next $\binom{n}{4}{a}^{n-4}{b}^{4}$
- In other words
- Thus, the very last term will be ${b}^{n} \,$
- The second last term will be $\binom{n}{n-1}a{b}^{n-1}$

Examples

${\left(a+b \right)}^{5}$
- Start with ${a}^{5} \,$
- The next term $\binom{5}{1}$ ${a}^{4} \,$ $b \,$
- And the next $\binom{5}{2}$ ${a}^{3} \,$ ${b}^{2} \,$
- Continue $\binom{5}{3}$ ${a}^{2} \,$ ${b}^{3} \,$
- Continue $\binom{5}{4}$ $a \,$ ${b}^{4} \,$
- Thus ${\left(a+b \right)}^{5}$ $={a}^{5}+\binom{5}{1}{a}^{4}b+\binom{5}{2}{a}^{3}{b}^{2}+\binom{5}{3}{a}^{2}{b}^{3}+\binom{5}{4}a{b}^{4}+ {b}^{5}$
- $={a}^{5}+5{a}^{4}b+10{a}^{3}{b}^{2}+10{a}^{2}{b}^{3}b+5a{b}^{4}+ {b}^{5} \,$

${\left(a+b \right)}^{3}$
Answer :
${\left(a+b \right)}^{3}={a}^{3}+\binom{3}{1}{a}^{2}b+\binom{3}{2}a{b}^{2}+{b}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^{3}$

Minus

What about ${\left(a-b \right)}^{n}$ ? Would we need another formula?

- Of course not. If the formula does not fit our needs
  - we DON'T change the formula
- So since the formula require a plus, we change it to a plus
  - ${\left(a-b \right)}^{n}$ $={\left[a+\left(?\right)\right]}^{n}$
  - It will be ${\left(a-b \right)}^{n}={\left[a+\left(-b\right)\right]}^{n}$
  - Now, in place of $b \,$ , we will put $\left(-b\right)$

${\left(a-b \right)}^{4}$
Take note
- ${\left(a-b \right)}^{4}$ $={a}^{4}+ \binom{4}{1}{a}^{3}{\color {Red}\left(-b\right)} +\binom{4}{2}{a}^{2}{{\color {Red}\left(-b\right)}}^{2}+\binom{4}{3}a{{\color {Red}\left(-b\right)}}^{3}+{{\color {Red}\left(-b\right)}}^{4}$
- $={a}^{4}-4{a}^{3}b+6{a}^{2}{b}^{2}-4a{b}^{3}+{b}^{4} \,$

Of Apples & Oranges

Actually, since there are three variables, we should call this apples & oranges & bananas :p, but we will leave the $n \,$ alone for now and concentrate on $a \,$ & $b \,$ .

The formula should then read

${\left({\color{Red}\heartsuit}+{\color{Blue}\bigcirc} \right)}^{n}$ $={{\color{Red}\heartsuit}}^{n}+\binom{n}{1}{{\color{Red}\heartsuit}}^{n-1}{\color{Blue}\bigcirc}+\binom{n}{2}{{\color{Red}\heartsuit}}^{n-2}{{\color{Blue}\bigcirc}}^{2}+\dots+\binom{n}{n-1}{\color{Red}\heartsuit}{{\color{Blue}\bigcirc}}^{n-1}+{{\color{Blue}\bigcirc}}^{n}$

Let's say we have

${\left(3x+2y \right)}^{4}$ So now the "apple" is ${\color{Red}3x}$ and the "orange" is ${\color{Blue}2y}$

Brackets

ALWAYS, I repeat, ALWAYS put a bracket around any non-single variable (including single negative variable such as $-b\,$ ) that we are going to use as the "apple" and "orange" EVERY single expansion term we write

Thus, when we write out the expansion for , we must see
- ${\left(3x+2y \right)}^{4}$ $={{\color{Red}\left(3x\right)}}^{4}+\binom{4}{1}{{\color{Red}\left(3x\right)}}^{3}{\color{Blue}\left(2y\right)}+\binom{4}{2}{{\color{Red}\left(3x\right)}}^{2}{{\color{Blue}\left(2y\right)}}^{2}+\binom{4}{3}{\color{Red}\left(3x\right)}{{\color{Blue}\left(2y\right)}}^{3}+{{\color{Blue}\left(2y\right)}}^{4}$

Common Mistakes

Forgetting to write the brackets i.e. $3{x}^4+ \binom{4}{1}3{x}^3\left(2y\right)+\binom{4}{2}3{x}^2 2{y}^2+ \dots$ is ABSOLUTELY WRONG
Putting the degree at wrongly

When to Simplify

NEVER, I repeat, NEVER EVER do any simplifying on the first step, or you will run into a high risk of making careless mistakes

How to Simplify

${\left(3x+2y \right)}^{4}$ $={\left(3x\right)}^{4}+\binom{4}{1}{\left(3x\right)}^{3}\left(2y\right)+\binom{4}{2}{\left(3x\right)}^{2}{\left(2y\right)}^{2}$ $+\binom{4}{3}\left(3x\right){\left(2y\right)}^{3}+{\left(2y\right)}^{4}$
So the next step is of course to simplify the terms
Remember that we do only write ONE single line of working. Writing more steps actually increases likelihood of careless mistakes.
Try the rest. The answer is $=81{x}^{4}+216{x}^{3}y+216{x}^{2}{y}^{2}+96x{y}^{3}+16{y}^{4} \,$

Sign

A very common mistake is writing the wrong sign (positive/negative) of the terms when there is a minus.

Firstly, remember that we MUST never attempt to determine the sign of the terms in the first step. Always keep the negative sign in the brackets
${\left(a-2b \right)}^{5}$
- $={a}^{5}-10{a}^{4}b+40{a}^{3}{b}^{2}-80{a}^{2}{b}^{3}+80a{b}^{4}-32{b}^{5} \,$

Notice the pattern?

Most of the time (except when it involves complex numbers), the pattern is either

all positive
all negative (except first term)
positive and negative alternately

Thus, if we get something like $+ \ldots - \ldots - \ldots + \ldots$ , there MUST be something wrong with the signs.

More Examples

Let's try a more challenging expansion. Remember all the steps explained above. Never try to take shortcuts and don't rush through things.

${\left({x}^{3}-\frac{2}{{x}^{2}} \right)}^{3}$
Try writing out the expansion
- Now try simplifying each term carefully

Now let's try something slightly different:

${\left( 1 + \sqrt{2}\right)}^{5}$
Notice that there are no unknowns, perhaps we can simplify the terms more than we normally could
- Try expanding it $={1}^{5}+\binom {5}{1}{1}^{4}\left( \sqrt{2}\right)+\binom {5}{2}{1}^{3}{\left( \sqrt{2}\right)}^{2}+\binom {5}{3}{1}^{2}{\left( \sqrt{2}\right)}^{3}+\binom {5}{4}1{\left( \sqrt{2}\right)}^{4}+{\left( \sqrt{2}\right)}^{5}$
- However,
- $=1+\binom {5}{1}\left( \sqrt{2}\right)+\binom {5}{2}{\left( \sqrt{2}\right)}^{2}+\binom {5}{3}{\left( \sqrt{2}\right)}^{3}+\binom {5}{4}{\left( \sqrt{2}\right)}^{4}+{\left( \sqrt{2}\right)}^{5}$
- Normally,
  - $=1+5\left( \sqrt{2}\right)+10\left(2\right)+10\left( 2\sqrt{2}\right)+5\left(4\right)+\left(4\sqrt{2}\right)$
  - $=41+29\sqrt{2}$
- Check quickly by

Now try this:

${\left( 1 - 2i \right)}^{4}$
Answer : $-7 + 24i \,$
Working:
- $=1-8i+24\left(-1 \right)-32\left(-i \right)+16$

Exercise 1

Expand and simplify the following

1) ${\left(x+y \right)}^{6}$ $={x}^{6}+6{x}^{5}y +15{x}^{4}{y}^{2}+20{x}^{3}{y}^{3}+15{x}^{2}{y}^{4}+6x{y}^{5}+{y}^{6} \,$

$\begin{align} {\left(x+y \right)}^{6} & = {x}^{6}+\binom{6}{1}{x}^{5}y +\binom{6}{2}{x}^{4}{y}^{2}+\binom{6}{3}{x}^{3}{y}^{3}+\binom{6}{4}{x}^{2}{y}^{4}+\binom{6}{5}x{y}^{5}+{y}^{6} \\ & = {x}^{6}+6{x}^{5}y +15{x}^{4}{y}^{2}+20{x}^{3}{y}^{3}+15{x}^{2}{y}^{4}+6x{y}^{5}+{y}^{6} \end{align}$

2) ${\left(x-y \right)}^{5}$ $={x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5} \,$

$\begin{align} {\left(x-y \right)}^{5} & = {x}^{5}+\binom{5}{1}{x}^{4}\left(-y\right)+\binom{5}{2}{x}^{3}{\left(-y\right)}^{2}+\binom{5}{3}{x}^{2}{\left(-y\right)}^{3}+ \binom{5}{4}x{\left(-y\right)}^{4}+{\left(-y\right)}^{5}\\ & = {x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5} \end{align}$

3) ${\left(3a-4b \right)}^{4}$ $={81a}^{4}-432{a}^{3}b+864{a}^{2}{b}^{2}-768a{b}^{3}+256{b}^{4} \,$

$\begin{align} {\left(3a-4b \right)}^{4} & = {\left(3a\right)}^{4}+\binom {4}{1}{\left(3a\right)}^{3}\left(-4b\right)+\binom {4}{2}{\left(3a\right)}^{2}{\left(-4b\right)}^{2} +\binom {4}{3}\left(3a\right){\left(-4b\right)}^{3}+{\left(-4b\right)}^{4}\\ & ={81a}^{4}-432{a}^{3}b+864{a}^{2}{b}^{2}-768a{b}^{3}+256{b}^{4} \end{align}$

4) ${\left(1+3{b}^{2}\right)}^{4}$ $=1+12{b}^{2}+54{b}^{4}+108{b}^{6}+81{b}^{8}\,$

$\begin{align} {\left(1+3{b}^{2}\right)}^{4} & = 1+\binom {4}{1}\left(3{b}^{2}\right)+\binom {4}{2}{\left( 3{b}^{2}\right)}^{2} +\binom {4}{3}{\left( 3{b}^{2}\right)}^{3}+{\left( 3{b}^{2}\right)}^{4}\\ & =1+12{b}^{2}+54{b}^{4}+108{b}^{6}+81{b}^{8} \end{align}$

5) ${\left(x+\frac{1}{x} \right)}^{5}$ $={x}^{5}+5{x}^{3}+10x+\frac{10}{x}+\frac{5}{{x}^{3}}+\frac{1}{{x}^{5}}$

$\begin{align} {\left(x+\frac{1}{x} \right)}^{5} & = {x}^{5}+\binom{5}{1}{x}^{4}\left(\frac{1}{x}\right)+\binom{5}{2}{x}^{3}{\left(\frac{1}{x}\right)}^{2} +\binom{5}{3}{x}^{2}{\left(\frac{1}{x}\right)}^{3}+\binom{5}{4}x{\left(\frac{1}{x}\right)}^{4}+{\left(\frac{1}{x}\right)}^{5} \\ & = {x}^{5}+5{x}^{3}+10x+\frac{10}{x}+\frac{5}{{x}^{3}}+\frac{1}{{x}^{5}} \end{align}$

6) ${\left({x}^{2}-\frac{1}{2{x}^{3}}\right)}^{7}$ $={x}^{14}-\frac{7}{2}{x}^{9}+\frac{21}{4}{x}^{4}-\frac{35}{8x}+\frac{35}{16{x}^{6}}-\frac{21}{32{x}^{11}}+\frac{7}{64{x}^{16}}-\frac{1}{128{x}^{21}}$

$\begin{align} {\left({x}^{2}-\frac{1}{2{x}^{3}}\right)}^{7} & = {\left({x}^{2}\right)}^{7}+\binom{7}{1}{\left({x}^{2}\right)}^{6}\left(-\frac{1}{2{x}^{3}}\right) +\binom{7}{2}{\left({x}^{2}\right)}^{5}{\left(-\frac{1}{2{x}^{3}}\right)}^{2} +\binom{7}{3}{\left({x}^{2}\right)}^{4}{\left(-\frac{1}{2{x}^{3}}\right)}^{3} +\binom{7}{4}{\left({x}^{2}\right)}^{3}{\left(-\frac{1}{2{x}^{3}}\right)}^{4} +\binom{7}{5}{\left({x}^{2}\right)}^{2}{\left(-\frac{1}{2{x}^{3}}\right)}^{5} +\binom{7}{6}\left({x}^{2}\right){\left(-\frac{1}{2{x}^{3}}\right)}^{6} +{\left(-\frac{1}{2{x}^{3}}\right)}^{7}\\ & = {x}^{14}-\frac{7}{2}{x}^{9}+\frac{21}{4}{x}^{4}-\frac{35}{8x}+\frac{35}{16{x}^{6}}-\frac{21}{32{x}^{11}}+\frac{7}{64{x}^{16}}-\frac{1}{128{x}^{21}} \end{align}$

7) ${\left(1+\sqrt{3}\right)}^{6}$ $= 208 + 120 \sqrt{3}$

$\begin{align} {\left(1+\sqrt{3}\right)}^{6} & = 1+\binom{6}{1}\sqrt{3} +\binom{6}{2}{\left(\sqrt{3}\right)}^{2}+\binom{6}{3}{\left(\sqrt{3}\right)}^{3} +\binom{6}{4}{\left(\sqrt{3}\right)}^{4}+\binom{6}{5}{\left(\sqrt{3}\right)}^{5}+{\left(\sqrt{3}\right)}^{6} \\ & = 1 + 6\sqrt{3}+15\left(3 \right)+20\left(3\sqrt{3} \right)+15\left(9 \right)+6\left(9\sqrt{3} \right)+27 \\ & = 208 + 120 \sqrt{3} \end{align}$

8) ${\left(3-2i\right)}^{4}$ $=-119-120i \,$

$\begin{align} {\left(3-2i\right)}^{4} & = {3}^{4}+\binom{4}{1}{3}^{3}\left(-2i\right)+\binom{4}{2}{3}^{2}{\left(-2i\right)}^{2}+\binom{4}{3}3{\left(-2i\right)}^{3}+{\left(-2i\right)}^{4} \\ & = 81 -216i+216{i}^{2}-96{i}^{3}+16{i}^{4} \\ & = 81 -216i-216-96\left(-i \right)+16 \\ & = -119-120i \end{align}$

9) ${\left(x+y\right)}^{5}-{\left(x-y\right)}^{5}$ . Hence, find the value of ${\left(\sqrt{2}+1\right)}^{5}-{\left(\sqrt{2}-1\right)}^{5}$ $10{x}^{4}y+20{x}^{2}{y}^{3}+2{y}^{5} ;82 \,$

- $\begin{align} {\left(x+y\right)}^{5}-{\left(x-y\right)}^{5} & = \left[{x}^{5}+\binom{5}{1}{x}^{4}y+\binom{5}{2}{x}^{3}{y}^{2}+\binom{5}{3}{x}^{2}{y}^{3}+\binom{5}{4}x{y}^{4}+{y}^{5}\right]- \left[{x}^{5}+\binom{5}{1}{x}^{4}\left(-y\right)+\binom{5}{2}{x}^{3}{\left(-y\right)}^{2}+\binom{5}{3}{x}^{2}{\left(-y\right)}^{3}+\binom{5}{4}x{\left(-y\right)}^{4}+{\left(-y\right)}^{5}\right]\\ & = \left[{x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\right]- \left[{x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}\right]\\ & = 10{x}^{4}y+20{x}^{2}{y}^{3}+2{y}^{5} \end{align}$
- ,
  - $\therefore {\left(\sqrt{2}+1\right)}^{5}-{\left(\sqrt{2}-1\right)}^{5}=10{\left(\sqrt{2}\right)}^{4}\left(1\right)+20{\left(\sqrt{2}\right)}^{2}{\left(1\right)}^{3}+2{\left(1\right)}^{5} =40+40+2=82$

The(r+1)th term

Sometimes we just need a particular term and not the whole expansion. Let's look at the formula again and try to see the pattern of each term

${\left(a+b \right)}^{n}$ $={a}^{n}+\binom{n}{1}{a}^{n-1}b+\binom{n}{2}{a}^{n-2}{b}^{2}+\binom{n}{3}{a}^{n-3}{b}^{3}+ \ldots$ $+\binom{n}{n-2}{a}^{2}{b}^{n-2}+\binom{n}{n-1}a{b}^{n-1}+{b}^{n}$
- Now look at the form of each term (except the first and the last),
- For example, lets look at $\binom{n}{2}{a}^{n-2}{b}^{2}$ . Notice where are the $2 \,$ 's $\binom{n}{{\color{Red}2}}{a}^{n-{\color{Red}2}}{b}^{{\color{Red}2}}$
- Similarly, for $\binom{n}{3}{a}^{n-3}{b}^{3}$ , $\binom{n}{{\color{Red}3}}{a}^{n-{\color{Red}3}}{b}^{{\color{Red}3}}$
- Thus, if we have a term that starts with $\binom {n}{10}$ , we know that the term will be $\binom{n}{10}{a}^{n-10}{b}^{10}$
- Also, for term with $\binom {n}{n-1}$ , we know that the term will be $\binom{n}{{\color{Red}n-1}}{a}^{n-{\color{Red}\left(n-1\right)}}{b}^{{\color{Red}n-1}}$ $=\binom{n}{n-1}a{b}^{n-1}$
- What about the first and the last term? We have seen that all the other terms have a form of
  - $\binom{n}{{\color{Red}\heartsuit}}{a}^{n-{\color{Red}\heartsuit}}{b}^{{\color{Red}\heartsuit}}$
- Thus, for the first term, which is , can we write in the same form?
  - ${a}^{n}= \,$ $\binom{n}{{\color{Red}?}}{a}^{n-{\color{Red}?}}{b}^{{\color{Red}?}}$
  - Thus we rewrite ${a}^{n}= \,$ $\binom{n}{0}{a}^{n-0}{b}^{0}$ . Note that $\binom {n}{0}=1$ and ${b}^{0}=1 \,$
- Notice how it fits nicely into the formula
- - Note that we have also rewritten $b \,=$ ${b}^{1} \,$
- What about the last term, ?
  - ${b}^{n}= \,$ $\binom{n}{{\color{Red}n}}{a}^{n-{\color{Red}n}}{b}^{{\color{Red}n}}$ . Note that $\binom {n}{n}=1$ and ${a}^{n-n}={a}^{0}=1 \,$
- Thus, the formula know looks like
  - ${\left(a+b \right)}^{n}=\binom{n}{0}{a}^{n-0}{b}^{0}+\binom{n}{1}{a}^{n-1}{b}^{1}+\binom{n}{2}{a}^{n-2}{b}^{2}+\binom{n}{3}{a}^{n-3}{b}^{3}+ \ldots+\binom{n}{n-2}{a}^{2}{b}^{n-2}+\binom{n}{n-1}{a}^{1}{b}^{n-1}+\binom{n}{n}{a}^{0}{b}^{n}$
- Thus, EVERY term now has the form of
  - $\binom{n}{{\color{Red}\heartsuit}}{a}^{n-{\color{Red}\heartsuit}}{b}^{{\color{Red}\heartsuit}}$ . We would use $r \,$ to instead of ${\color{Red}\heartsuit}$
- Thus $\binom{n}{{\color{Red}r}}{a}^{n-{\color{Red}r}}{b}^{{\color{Red}r}}$
- Important : Where $r \,$ takes values from $0 \,$ to $n \,$ and NOT $1 \,$ to $n \,$

But how would we know what value of $r \,$ to use for which term?

Let's name the terms first (according to position)
- ${\left(a+b \right)}^{n}= \overbrace{\binom{n}{0}{a}^{n-0}{b}^{0}}^{{1}^{st} term} +\overbrace{\binom{n}{1}{a}^{n-1}{b}^{1}}^{{2}^{nd} term} +\overbrace{\binom{n}{2}{a}^{n-2}{b}^{2}}^{{3}^{rd} term} +\overbrace{\binom{n}{3}{a}^{n-3}{b}^{3}}^{{4}^{th} term} +\ldots +\overbrace{\binom{n}{n-2}{a}^{2}{b}^{n-2}}^{{\left(n-1\right)}^{th} term} +\overbrace{\binom{n}{n-1}{a}^{1}{b}^{n-1}}^{{n}^{th} term} +\overbrace{\binom{n}{n}{a}^{0}{b}^{n}}^{{\left(n+1\right)}^{th} term}$
- Note :
- But the $r \,$ does not start from $1 \,$ ,
- ${\left(a+b \right)}^{n}= \underbrace{\overbrace{\binom{n}{0}{a}^{n-0}{b}^{0}}^{{1}^{st} term}}_{r=?} +\underbrace{\overbrace{\binom{n}{1}{a}^{n-1}{b}^{1}}^{{2}^{nd} term}}_{r=?} +\underbrace{\overbrace{\binom{n}{2}{a}^{n-2}{b}^{2}}^{{3}^{rd} term}}_{r=?} +\underbrace{\overbrace{\binom{n}{3}{a}^{n-3}{b}^{3}}^{{4}^{th} term}}_{r=?} +\ldots +\underbrace{\overbrace{\binom{n}{n-2}{a}^{2}{b}^{n-2}}^{{\left(n-1\right)}^{th} term}}_{r=?} +\underbrace{\overbrace{\binom{n}{n-1}{a}^{1}{b}^{n-1}}^{{n}^{th} term}}_{r=?} +\underbrace{\overbrace{\binom{n}{n}{a}^{0}{b}^{n}}^{{\left(n+1\right)}^{th} term}}_{r=?}$
- ${\left(a+b \right)}^{n}= \underbrace{\overbrace{\binom{n}{0}{a}^{n-0}{b}^{0}}^{{1}^{st} term}}_{r=0} +\underbrace{\overbrace{\binom{n}{1}{a}^{n-1}{b}^{1}}^{{2}^{nd} term}}_{r=1} +\underbrace{\overbrace{\binom{n}{2}{a}^{n-2}{b}^{2}}^{{3}^{rd} term}}_{r=2} +\underbrace{\overbrace{\binom{n}{3}{a}^{n-3}{b}^{3}}^{{4}^{th} term}}_{r=3} +\ldots +\underbrace{\overbrace{\binom{n}{n-2}{a}^{2}{b}^{n-2}}^{{\left(n-1\right)}^{th} term}}_{r=n-2} +\underbrace{\overbrace{\binom{n}{n-1}{a}^{1}{b}^{n-1}}^{{n}^{th} term}}_{r=n-1} +\underbrace{\overbrace{\binom{n}{n}{a}^{0}{b}^{n}}^{{\left(n+1\right)}^{th} term}}_{r=n}$
- Thus, we can conclude, for the ${10}^{th} \,$ term, for example, $r= \,$ $9 \,$
- On the other hand, if $r = 7 \,$ , it represents the ${8}^{th} \,$ term.
- So which term does represent? ${\left(r+1\right)}^{th}$ term.
  - ${\left(a+b \right)}^{n}= \underbrace{\overbrace{\binom{n}{0}{a}^{n-0}{b}^{0}}^{{1}^{st} term}}_{r=0} +\underbrace{\overbrace{\binom{n}{1}{a}^{n-1}{b}^{1}}^{{2}^{nd} term}}_{r=1} +\underbrace{\overbrace{\binom{n}{2}{a}^{n-2}{b}^{2}}^{{3}^{rd} term}}_{r=2} +\ldots +\underbrace{\overbrace{\binom{n}{r}{a}^{n-r}{b}^{r}}^{{r+1}^{th} term}}_{r=r} +\ldots +\underbrace{\overbrace{\binom{n}{n-2}{a}^{2}{b}^{n-2}}^{{\left(n-1\right)}^{th} term}}_{r=n-2} +\underbrace{\overbrace{\binom{n}{n-1}{a}^{1}{b}^{n-1}}^{{n}^{th} term}}_{r=n-1} +\underbrace{\overbrace{\binom{n}{n}{a}^{0}{b}^{n}}^{{\left(n+1\right)}^{th} term}}_{r=n}$

Thus, the formula is

${r+1}^{th} \,$ term, ${T}_{r+1}=\binom{n}{r}{a}^{n-r}{b}^{r}$

We can also use $\sum$ notation to represent the whole expansion

${\left(a+b \right)}^{n}=$ $\sum_{r=0}^n \binom{n}{r}{a}^{n-r}{b}^{r}$ . Note the $r=0 \,$

When to Use the formula

First, before we even start to see how to use the formula, we must know WHEN to use the formula.

What does the formula gives us?
- A specific term.
- Thus, it allows us to
- In other words, we would use this formula when the questions asks for

Using the formula

First step, would be memorizing the formula

${T}_{r+1}= \,$ $\binom{n}{r}{a}^{n-r}{b}^{r}$

The next step, would be to substitute $a,b,n \,$ from the wanted expansion. This actually takes a bit of practice.

- ${T}_{r+1}\,$ $=\binom{{\color{Red}7}}{r}{x}^{{\color{Red}7}-r}{y}^{r}$
. Becareful of "apples" and "oranges"!
- ${T}_{r+1}\,$ $=\binom{4}{r}{{\color{Red}\left(2a\right)}}^{4-r}{{\color{Blue}\left(-b\right)}}^{r}$
- ${T}_{r+1}\,$ $=\binom{{\color{Red}2n}}{r}{a}^{{\color{Red}2n}-r}{b}^{r}$

Determining Value of r

The next, and the most tricky step, is determining the value of $r \,$ needed for the term required.

Again, first remember the formula is ${T}_{r+1}=\binom{n}{r}{a}^{n-r}{b}^{r}$ . In other words, when we use $r \,$ , it represents ${\left(r+1\right)}^{th}$ term.

Example

In the expansion of ${\left(3+2x\right)}^{10}$ , find
${6}^{th} \,$ term
Before determining value of needed, let's write the formula first
- ${T}_{r+1}\,$ $=\binom{10}{r}{3}^{10-r}{\left(2x\right)}^{r}$
- $\mbox{For }{T}_{6}\,$ , $r= \,$
- $\therefore {T}_{6}$ $=\binom{10}{5}{3}^{10-5}{\left(2x\right)}^5$ $=1959552{x}^{5} \,$

${8}^{th} \,$ term
- $\mbox{For }{T}_{8}\,$ , $r=7 \,$ . ${T}_{8}=\binom{10}{7}{3}^{3}{\left(2x\right)}^7=414720{x}^{7}$

term in ${x}^{3} \,$
- This would be a more challenging question. Make sure you understand the steps that we take here.

term in ${x}^{5} \,$
- ${T}_{r+1}=\binom{10}{r}{3}^{10-r}{\left(2x\right)}^{r}$ . $\mbox{For term in }{x}^{5}, r=5 \,$ . $\therefore \mbox{term in }{x}^{5}=\binom{10}{5}{3}^{5}{\left(2x\right)}^{5}=1959552{x}^{5}$

IMPORTANT :

In the expansion of ${\left({x}^{3}-\frac{2}{{x}^{2}} \right)}^{5}$ , find
term in ${x}^{5} \,$
- Firstly, $=\binom{5}{r}{\left({x}^{3}\right)}^{5-r}{\left(-\frac{2}{{x}^{2}}\right)}^{r}$
  - Now, what value of $r \,$ would we use to obtain ${x}^{5} \,$ ?
- Thus we now have
- $\mbox{For term in }{x}^{5} \,$
- $15-5r=5 \,$
- $\therefore r=2 \,$

constant term / term independent of $x \,$
- Note :
- - $15-5r=0 \,$
  - $r=3 \,$
  - $\therefore \mbox{constant term }$ $=\binom{5}{3}{\left({x}^{3}\right)}^{2}{\left(-\frac{2}{{x}^{2}}\right)}^{3}$ $=-80\,$

coefficient of the term in $\frac{1}{{x}^{5}}$
- $\mbox{For term in }\frac{1}{{x}^{5}}$
- $15-5r=-5 \,$
- $r=4 \,$
- $\therefore \mbox{term in }\frac{1}{{x}^{5}}$ $=\binom{5}{4}{\left({x}^{3}\right)}^{1}{\left(-\frac{2}{{x}^{2}}\right)}^{4}$ $=\frac{80}{{x}^{5}}$
- Correct? You sure?

More Examples

- ${T}_{r+1}= \,$ $\binom{7}{r}{1}^{7-r}{\left(3{x}^{2}\right)}^{r}$
- $\mbox{Degree of }x =\,$ $2r \,$

- ${T}_{r+1}= \,$ $\binom{5}{r}{\left(3x\right)}^{5-r}{\left(-\frac{2}{x}\right)}^{r}$
- $\mbox{Degree of }x =\,$ $\left(5-r\right)-r \,$

- ${T}_{r+1}= \,$ $\binom{6}{r}{\left({x}^{2}\right)}^{6-r}{\left(\frac{1}{2{x}^{3}}\right)}^{r}$
- $\mbox{Degree of }x =\,$ $2\left(6-r\right)-3r \,$

Exercise 2

In the following expansions, find the terms/coefficients stated

1) ${\left(3-2x \right)}^{9}$ ; ${3}^{rd} \,$ and ${6}^{th} \,$ term $314928{x}^{2} ; -326592{x}^{5} \,$

$\begin{align} & {T}_{r+1}=\binom{9}{r}{3}^{9-r}{\left(-2x\right)}^{r} \\ & \mbox{For }{3}^{rd} \mbox{ term, }r=2 \\ & {T}_{3}=\binom{9}{2}{3}^{7}{\left(-2x\right)}^{2}=314928{x}^{2}\\ & \mbox{For }{6}^{th} \mbox{ term, }r=5 \\ & {T}_{6}=\binom{9}{5}{3}^{4}{\left(-2x\right)}^{5}=-326592{x}^{5}\\ \end{align}$

2) ${\left(1+x \right)}^{15}$ ; term in ${x}^{5} \,$ and term in ${x}^{12} \,$ $3003{x}^{5} ; 455{x}^{12} \,$

$\begin{align} & {T}_{r+1}=\binom{15}{r}{x}^{r} \\ & \mbox{For term in }{x}^{5}, r=5 \\ & \therefore \mbox{term in }{x}^{5}=\binom{15}{5}{x}^{5}=3003{x}^{5}\\ & \mbox{For term in }{x}^{12}, r=12 \\ & \therefore \mbox{term in }{x}^{12}=\binom{15}{12}{x}^{12}=455{x}^{12}\\ \end{align}$

3) ${\left(x+2y \right)}^{8}$ ; term in ${x}^{2}{y}^{6} \,$ $=1792{x}^{2}{y}^{6} \,$

$\begin{align} & {T}_{r+1}=\binom{8}{r}{x}^{8-r}{\left(2y\right)}^{r} \\ & \mbox{For term in }{x}^{2}{y}^{6}, r=6 \\ & \therefore \mbox{term in }{x}^{2}{y}^{6}=\binom{8}{6}{x}^{2}{\left(2y\right)}^{6}=1792{x}^{2}{y}^{6}\\ \end{align}$

4) ${\left(1-3{x}^{2} \right)}^{10}$ ; coefficient of term in ${x}^{6}\,$ and coefficient of term in ${x}^{8}\,$ $-3240 ; 17010\,$

$\begin{align} & {T}_{r+1}=\binom{10}{r}{\left(-3{x}^{2}\right)}^{r} \\ & \mbox{Degree of }x =2r \\ & \mbox{For term in }{x}^{6}, 2r=6, \therefore r=3 \\ & \therefore \mbox{term in }{x}^{6}=\binom{10}{3}{\left(-3{x}^{2}\right)}^{3}=-3240{x}^{6}\\ & \mbox{coefficient of term in }{x}^{6}=-3240 \\ & \mbox{For term in }{x}^{8}, 2r=8, \therefore r=4 \\ & \therefore \mbox{term in }{x}^{8}=\binom{10}{4}{\left(-3{x}^{2}\right)}^{4}=17010{x}^{8}\\ & \mbox{coefficient of term in }{x}^{8}=17010 \\ \end{align}$

5) ${\left(x+\frac{1}{x}\right)}^{8}$ ; constant term and term in $\frac{1}{{x}^{4}}\,$ $70 ; \frac{28}{{x}^{4}}$

$\begin{align} & {T}_{r+1}=\binom{8}{r}{x}^{8-r}{\left(\frac{1}{x}\right)}^{r} \\ & \mbox{Degree of }x =8-r-r=8-2r \\ & \mbox{For constant term }, 8-2r=0, \therefore r=4\\ & \therefore \mbox{constant term }=\binom{8}{4}{x}^{4}{\left(\frac{1}{x}\right)}^{4}=70\\ & \mbox{For term in }\frac{1}{{x}^{4}}, 8-2r=-4, \therefore r=6 \\ & \therefore \mbox{term in }\frac{1}{{x}^{4}}=\binom{8}{6}{x}^{2}{\left(\frac{1}{x}\right)}^{6}=\frac{28}{{x}^{4}}\\ \end{align}$

6) ${\left(2{x}^{3}-\frac{1}{3{x}^{2}}\right)}^{9}$ ; term in ${x}^{7}\,$ and term in $\frac{1}{{x}^{8}}\,$ $\frac{448}{9}{x}^{7} ; -\frac{16}{243{x}^{8}}$

$\begin{align} & {T}_{r+1}=\binom{9}{r}{\left(2{x}^{3}\right)}^{9-r}{\left(-\frac{1}{3{x}^{2}}\right)}^{r} \\ & \mbox{Degree of }x= 3\left(9-r \right)-2r=27-5r\, \\ & \mbox{For term in }{x}^{7}, 27-5r=7, \therefore r=4\\ & \therefore \mbox{term in }{x}^{7}=\binom{9}{4}{\left(2{x}^{3}\right)}^{5}{\left(-\frac{1}{3{x}^{2}}\right)}^{4}=\frac{448}{9}{x}^{7}\\ & \mbox{For term in }\frac{1}{{x}^{8}}, 27-5r=-8, \therefore r=7\\ & \therefore \mbox{term in }\frac{1}{{x}^{8}}=\binom{9}{7}{\left(2{x}^{3}\right)}^{2}{\left(-\frac{1}{3{x}^{2}}\right)}^{7}=-\frac{16}{243{x}^{8}}\\ \end{align}$

Ascending/Descending Powers

${\left(a+b \right)}^{n}$ $={a}^{n}+\binom{n}{1}{a}^{n-1}b+\binom{n}{2}{a}^{n-2}{b}^{2}+\dots+\binom{n}{n-1}a{b}^{n-1}+{b}^{n}$

The powers of $b \,$ are ascending(increasing)
The powers of $a \,$ are descending(decreasing)

If, however, we need it to be the other way round, we can exchange their positions

${\left(a+b \right)}^{n}=$ ${\left(b+a \right)}^{n}=$ ${b}^{n}+\binom{n}{1}{b}^{n-1}a+\binom{n}{2}{b}^{n-2}{a}^{2}+\dots+\binom{n}{n-1}b{a}^{n-1}+{a}^{n}$

The powers of $a \,$ are ascending
The powers of $b \,$ are descending

Formula for (1+x)^n

In the special case where $a=1, b=x \,$

- $={1}^{n}+\binom{n}{1}{1}^{n-1}x+\binom{n}{2}{1}^{n-2}{x}^{2}+\dots+{x}^{n}$
- $=1+\binom{n}{1}x+\binom{n}{2}{x}^{2}+\dots+{x}^{n}$

Expand until

When we are only required to expand only until a certain terms, there are a few ways the question can be written

Example : Expand the following in ascending powers powers of $x \,$

a) until the term in ${x}^{2} \,$
b) neglecting /omitting the term in ${x}^{3} \,$ and above
c) if the term in ${x}^{3} \,$ above are omitted
d) if $x\,$ is so small that $x^{3}\,$ and higher powers of $x\,$ can be ignored

Note that all the ways above have the same meaning, since if we expand till the term in ${x}^{2} \,$ only, it means we are omitting the terms in ${x}^{3} \,$ and above.

It is thus very important to read the question carefully, to note whether we are required to do ascending or descending, and to expand until where.

IMPORTANT : Since the expansion is not complete, we MUST leave $+\ldots$ at the end to show that we understand there are more terms behind.

Examples

Expand the following in ascending powers of $x \,$ until the term in ${x}^{3} \,$

${\left(1+x \right)}^{7}$
- Analyze :
- ${\left(1+x \right)}^{7}=1+\binom{7}{1}x+\binom{7}{2}{x}^{2}+\binom{7}{3}{x}^{3}+\ldots$
- $=1+7x+21{x}^{2}+35{x}^{3}+\ldots$

${\left(x+2 \right)}^{5}$
- Analyze :
- ${\left(x+2 \right)}^{5}$ $={\left(2+x \right)}^{5}$
- $={2}^{5}+\binom{5}{1}{2}^{4}x+\binom{5}{2}{2}^{3}{x}^{2}+\binom{5}{3}{2}^{2}{x}^{3}+\ldots$
- $=32+80x+80{x}^{2}+40{x}^{3}+\ldots$

${\left(1+{x}^{2} \right)}^{6}$
- $=1+6{x}^{2}+\ldots$

${\left(1+2x \right)}^{4}-{\left(1+3x \right)}^{5}$
- Analyze :
- ${\left(1+2x \right)}^{4}-{\left(1+3x \right)}^{5}$
- $=\left[1+\binom{4}{1}\left(2x \right)+\binom{4}{2}{\left(2x \right)}^{2}+\binom{4}{3}{\left(2x \right)}^{3}+\ldots\right]$ $-\left[1+\binom{5}{1}\left(3x \right)+\binom{5}{2}{\left(3x \right)}^{2}+\binom{5}{3}{\left(3x \right)}^{3}+\ldots\right]$
- $=\left(1+8x+24{x}^{2}+32{x}^{3}+\ldots \right)$ $-\left(1+15x+90{x}^{2}+270{x}^{3}+\ldots \right)$
- $=-7x-66{x}^{2}-238{x}^{3}+\ldots$

Multiplying with Polynomial/Another Expansion

Normally, we would only need the first few terms (in ascending powers), so

we rearrange (if needed) all expansions/polynomials in ascending powers
expand only until the required terms
multiply and simplify carefully

Example Expand the following in ascending powers of $x \,$ , neglecting the term ${x}^{3}\,$ and above.

We expand in ascending powers till ${x}^{2}\,$

$\left(x+2 \right){\left(1+2x \right)}^{4}$
- Analyze :
- $\left(x+2 \right){\left(1+2x \right)}^{4}$ $=\left(2+x \right){\left(1+2x \right)}^{4}$ Next, we expand the second bracket
- $=\left(2+x \right)\left[1+\binom{4}{1}\left(2x \right)+\binom{4}{2}{\left(2x \right)}^{2}+\ldots \right]$
- $=\left(2+x \right)\left(1+8x+24{x}^{2}+\ldots \right)$
- The next step is tricky. We
- $={\color{Red}2+16x+48{x}^{2}}+{\color{Blue}x+8{x}^{2}}+\ldots$ Notice we need to expand only till ${x}^{2} \,$
- $=2+17x+56{x}^{2}+\ldots$

$\left(1-{x}^{2} \right){\left(1+x \right)}^{5}$
Analyze : Already arranged nicely
- $=\left(1-{x}^{2} \right)\left[1+\binom{5}{1}x+\binom{5}{2}{x}^{2}+\ldots \right]$
- $=\left(1-{x}^{2} \right)\left(1+5x+10{x}^{2}+\ldots \right)$ $\left({\color{Red}1}-{\color{Blue}{x}^{2}} \right)\left(1+5x+10{x}^{2}+\ldots \right)$
- $={\color{Red}1+5x+10{x}^{2}}{\color{Blue}-{x}^{2}}+\ldots$
- $=1+5x+9{x}^{2}+\ldots$

${\left(1-x \right)}^{5}{\left(1+3x \right)}^{4}$
- Analyze :
- ${\left(1-x \right)}^{5}$ $=1+\binom{5}{1}\left(-x \right)+\binom{5}{2}{\left(-x \right)}^{2}+\ldots$ $=1-5x+10{x}^{2}+\ldots$
- ${\left(1+3x \right)}^{4}$ $=1+\binom{4}{1}\left(3x \right)+\binom{4}{2}{\left(3x \right)}^{2}+\ldots$ $=1+12x+54{x}^{2}+\ldots$
- $\therefore {\left(1-x \right)}^{5}{\left(1+3x \right)}^{4}$ $=\left(1-5x+10{x}^{2}+\ldots\right) \left(1+12x+54{x}^{2}+\ldots\right)$
- $={\color{Red}1+12x+54{x}^{2}}-5x-60{x}^{2}+{\color{Blue}10{x}^{2}}+\ldots$
- $=1+7x+4{x}^{2} \ldots$
- Note:

Expansion of (a+b+c)^n

We have the formula for ${\left(a+b \right)}^{n}$ , but what about ${\left(a+b+c \right)}^{n}$ ?

Again, we don't change the formula, we

The formula asks for two variable only, so we give it two variable

${\left(a+b+c \right)}^{n}=$ ${\left({\color{Red}\heartsuit}+{\color{Blue}\bigcirc} \right)}^{n}$

We can either do it by

${\left[{\color{Red}a}+{\color{Blue}\left( b+c\right)}\right]}^{n}$
${\left[{\color{Red}\left(a+b\right)}+{\color{Blue}c}\right]}^{n}$

Example ${\left(a+b+c \right)}^{3}$

$={\left[a+\left(b+c\right) \right]}^{3}$
$={a}^{3}+\binom {3}{1}{a}^{2}\left(b+c\right)+\binom {3}{2}a{\left(b+c\right)}^{2}+{\left(b+c\right)}^{3}$

Example : Expand the following in ascending powers of $x \,$ until the term in ${x}^{3} \,$

${\left(1+x+{x}^{2} \right)}^{4}$
- Analyze:
- $={\left[1+\left(x+{x}^{2}\right)\right]}^{4}$
- $=1+\binom{4}{1}\left(x+{x}^{2}\right)+\binom{4}{2}{\left(x+{x}^{2}\right)}^{2}+\binom{4}{3}{\left(x+{x}^{2}\right)}^{3}+\dots$
- - ${\left(x+{x}^{2}\right)}^{2}$ $={x}^{2}+2{x}^{3}+\ldots$ The last term is not needed
  - ${\left(x+{x}^{2}\right)}^{3}$ $={x}^{3}+\ldots$ All other terms have higher powers
- $=1+4\left(x+{x}^{2}\right)+6\left({x}^{2}+2{x}^{3}+\ldots \right)+4\left({x}^{3}+\ldots \right)+\ldots$
- $=1+4x+4{x}^{2}+6{x}^{2}+12{x}^{3}+4{x}^{3}+\ldots$
- $=1+4x+10{x}^{2}+16{x}^{3}+\ldots$

Exercise 3

Expand the following in ascending powers of $x \,$ until the term in ${x}^{3} \,$

1) $\left(1+x \right){\left(1-3x \right)}^{5}$ $=1-14x+75{x}^{2}-180{x}^{3}+\ldots$

$\begin{align} \left(1+x \right){\left(1-3x \right)}^{5} & = \left(1+x \right)\left[1+\binom{5}{1}\left(-3x\right)+\binom{5}{2}{\left(-3x\right)}^{2}+\binom{5}{3}{\left(-3x\right)}^{3}+\ldots\right]\\ & = \left(1+x \right)\left(1-15x+90{x}^{2}-270{x}^{3}+\ldots \right)\\ & = 1-15x+90{x}^{2}-270{x}^{3}+x-15{x}^{2}+90{x}^{3}+\ldots \\ & = 1-14x+75{x}^{2}-180{x}^{3}+\ldots \end{align}$

2) $\left({x}^{2}+3 \right){\left(1-2x \right)}^{6}$ $= 3-36x+181{x}^{2}-492{x}^{3}+\ldots$

$\begin{align} \left({x}^{2}+3 \right){\left(1-2x \right)}^{6} & = \left(3+{x}^{2}\right) \left[1+\binom{6}{1}\left(-2x\right)+\binom{6}{2}{\left(-2x\right)}^{2}+\binom{6}{3}{\left(-2x\right)}^{3}+\ldots\right]\\ & = \left(3+{x}^{2}\right) \left(1-12x+60{x}^{2}-160{x}^{3}+\ldots \right)\\ & = 3-36x+180{x}^{2}-480{x}^{3}+{x}^{2}-12{x}^{3}+\ldots\\ & = 3-36x+181{x}^{2}-492{x}^{3}+\ldots \end{align}$

3) $\left(1-x+{x}^{2} \right){\left(1+x \right)}^{7}$ $=1+6x+15{x}^{2}+21{x}^{3}+\ldots$

$\begin{align} \left(1-x+{x}^{2} \right){\left(1+x \right)}^{7} & = \left(1-x+{x}^{2} \right)\left[1+\binom{7}{1}x+\binom{7}{2}{x}^{2}+\binom{7}{3}{x}^{3}+\ldots \right]\\ & = \left(1-x+{x}^{2} \right)\left(1+7x+21{x}^{2}+35{x}^{3}+\ldots \right) \\ & = 1+7x+21{x}^{2}+35{x}^{3}-x-7{x}^{2}-21{x}^{3}+{x}^{2}+7{x}^{3}+\ldots \\ & = 1+6x+15{x}^{2}+21{x}^{3}+\ldots \end{align}$

4) ${\left( x+1\right)}^{4}{\left(2x+1 \right)}^{6}$ $=1+16x+114{x}^{2}+476{x}^{3}+\ldots$

- $\begin{align} {\left( x+1\right)}^{4} & = {\left( 1+x\right)}^{4}\\ & = \left[1+\binom{4}{1}x+\binom{4}{2}{x}^{2}+\binom{4}{3}{x}^{3}+\ldots \right]\\ & = 1+4x+6{x}^{2}+4{x}^{3}+\ldots \end{align}$
- $\begin{align} {\left( 2x+1\right)}^{6} & = {\left( 1+2x\right)}^{6}\\ & = \left[1+\binom{6}{1}\left(2x \right)+\binom{6}{2}{\left(2x \right)}^{2}+\binom{6}{3}{\left(2x \right)}^{3}+\ldots \right]\\ & = 1+12x+60{x}^{2}+160{x}^{3}+\ldots \end{align}$
- $\begin{align} \therefore {\left( x+1\right)}^{4}{\left(2x+1 \right)}^{6} & = \left(1+4x+6{x}^{2}+4{x}^{3}+\ldots\right)\left(1+12x+60{x}^{2}+160{x}^{3}+\ldots\right)\\ & = 1+12x+60{x}^{2}+160{x}^{3}+4x+48{x}^{2}+240{x}^{3}+6{x}^{2}+72{x}^{3}+4{x}^{3}+\ldots\\ & = 1+16x+114{x}^{2}+476{x}^{3}+\ldots \end{align}$

5) ${\left(1+x-{x}^{2} \right)}^{4}$ $= 1+4x+2{x}^{2}-8{x}^{3}+\ldots$

$\begin{align} {\left(1+x-{x}^{2} \right)}^{4} & = {\left[1+\left(x-{x}^{2}\right)\right]}^{4} \\ & = 1+\binom{4}{1}\left(x-{x}^{2}\right)+\binom{4}{2}{\left(x-{x}^{2}\right)}^{2}+\binom{4}{3}{\left(x-{x}^{2}\right)}^{3}+\ldots \\ & = 1+4\left(x-{x}^{2}\right)+6\left({x}^{2}-2{x}^{3}+\ldots \right)+4\left({x}^{3}+\ldots \right)+\ldots \\ & = 1+4x-4{x}^{2}+6{x}^{2}-12{x}^{3}+4{x}^{3}+\ldots \\ & = 1+4x+2{x}^{2}-8{x}^{3}+\ldots \end{align}$

6) ${\left(1-3x+2{x}^{2} \right)}^{6}$ $= 1-18x+147{x}^{2}-720{x}^{3}+\ldots$

$\begin{align} {\left(1-3x+2{x}^{2} \right)}^{6} & = {\left[1+\left(-3x+2{x}^{2}\right)\right]}^{6} \\ & = 1+\binom{6}{1}\left(-3x+2{x}^{2} \right)+\binom{6}{2}{\left(-3x+2{x}^{2} \right)}^{2}+\binom{6}{3}{\left(-3x+2{x}^{2} \right)}^{3}+\ldots \\ & = 1+6\left(-3x+2{x}^{2} \right)+15\left(9{x}^{2}-12{x}^{3}+\ldots \right)+20\left(-27{x}^{3}+\ldots \right)+\ldots \\ & = 1-18x+12{x}^{2}+135{x}^{2}-180{x}^{3}-540{x}^{3}+\ldots \\ & = 1-18x+147{x}^{2}-720{x}^{3}+\ldots \end{align}$

The(r+1)th term (Part 2)

To find certain terms from expansions that involve multiplication with other polynomial/expansions, we have to analyze the situation very carefully. The formula for ${T}_{r+1}\,$ helps, but ONLY after we analyze what terms we need.

Example 1 : Find the ${x}^{5}\,$ in the expansion $\left(3+x \right){\left(1+2x \right)}^{6}$

- Analyze:
- $\therefore \left(3+x \right){\left(1+2x \right)}^{6}$ $=\left(3+x \right)\left(\ldots+240{x}^{4}+192{x}^{5}+\ldots\right)$
- $\therefore \mbox{term in } {x}^{5}\,$ $=3\left(192{x}^{5}\right)+240{x}^{5}$ $=816{x}^{5}\,$
- Note :

Example 2 : Find the coefficient of the term in ${x}^{3}\,$ in the expansion $\left(1+x+{x}^{2} \right){\left(1+x \right)}^{8}$

- Analyze: $\left({\color{Red}1}+{\color{Blue}x}+{\color{Purple}{x}^{2}}\right)\left(1\ldots {\color{Purple}x} \dots {\color{Blue}{x}^{2}}\ldots {\color{Red}{x}^{3}}\ldots \right)$ Might as well expand like usual.
- $\left(1+x+{x}^{2}\right){\left(1+x \right)}^{8}$ $=\left(1+x+{x}^{2} \right)\left[1+\binom{8}{1}x+\binom{8}{2}{x}^{2}+\binom{8}{3}{x}^{3}+\ldots \right]$
- $=\left(1+x+{x}^{2} \right)\left(1+8x+28{x}^{2}+56{x}^{3}+\ldots \right)$
- $\therefore \mbox{term in }{x}^{3}\,$ $=56{x}^{3}+28{x}^{3}+8{x}^{3}=92{x}^{3}\,$
- $\mbox{coefficient of term in }{x}^{3}\,$ $=92\,$

Exercise 4

In the following expansions, find the terms/coefficients stated

1) $\left(1-2x\right){\left(1+x \right)}^{7}$ ; term in ${x}^{6}\,$ $-35{x}^{6}\,$

$\begin{align} &\left(1-2x\right){\left(1+x \right)}^{7}\\ &=\left(1-2x\right)\left[\ldots+\binom{7}{5}{x}^{5}+\binom{7}{6}{x}^{6}+\ldots\right]\\ &=\left(1-2x\right)\left(\ldots+21{x}^{5}+7{x}^{6}+\ldots\right)\\ & \mbox{term in }{x}^{6}=7{x}^{6}-42{x}^{6}=-35{x}^{6}\\ \end{align}$

2) $\left(1+{x}^{2}\right){\left(1-2x \right)}^{10}$ ; term in ${x}^{8}\,$ $24960{x}^{8}\,$

$\begin{align} &\left(1+{x}^{2}\right){\left(1-2x \right)}^{10}\\ &=\left(1+{x}^{2}\right)\left[\ldots+\binom{10}{6}{\left(-2x\right)}^{6}+\ldots+\binom{10}{8}{\left(-2x\right)}^{8}+\ldots\right]\\ &=\left(1+{x}^{2}\right)\left(\ldots+13440{x}^{6}+\ldots+11520{x}^{8}+\ldots\right)\\ & \mbox{term in }{x}^{8}=11520{x}^{8}+13440{x}^{8}=24960{x}^{8}\\ \end{align}$

3) $\left(1+2x+{x}^{2}\right){\left(1+{x}^{2}\right)}^{12}$ ; term in ${x}^{11}\,$ $1584{x}^{11}\,$

$\begin{align} &\left(1+2x+{x}^{2}\right){\left(1+{x}^{2}\right)}^{12}=\left(1+2x+{x}^{2}\right)\left[\ldots{x}^{6}\ldots{x}^{8}\ldots{x}^{10}\ldots\right]\\ & \mbox{For }{\left(1+{x}^{2}\right)}^{12},\;{T}_{r+1}=\binom{12}{r}{\left({x}^{2}\right)}^{r}\\ & \mbox{For term in }{x}^{10}, 2r=10, \therefore r=5\\ &\therefore \mbox{term in }{x}^{11}=\left(2x\right)\left[\binom{12}{5}{\left({x}^{2}\right)}^{5}\right]=1584{x}^{11}\\ \end{align}$

4) $x{\left(\frac{x}{2}+\frac{3}{x}\right)}^{10}$ ; coefficient of term in $x\,$ $\frac{15309}{8}$

$\begin{align} & \mbox{For }{\left(\frac{x}{2}+\frac{3}{x}\right)}^{10},\;{T}_{r+1}=\binom{10}{r}{\left(\frac{x}{2}\right)}^{10-r}{\left(\frac{3}{x}\right)}^{r}\\ &\mbox{Degree of }x=\left(10-r\right)-r=10-2r\\ &\mbox{For constant term }10-2r=0, \therefore r=5\\ &\therefore \mbox{term in }x=x\left[\binom{10}{5}{\left(\frac{x}{2}\right)}^{5}{\left(\frac{3}{x}\right)}^{5}\right]=\frac{15309}{8}x\\ &\mbox{coefficient of term in }x=\frac{15309}{8}\\ \end{align}$

Binomial Expansion Part1

From StpmWiki

Contents

Notes

Learning Objectives (Syllabus)

Prior Knowledge

Binomial Coefficient

Important values

Expansion for positive integer n

Examples

Minus

Of Apples & Oranges

Brackets

Common Mistakes

When to Simplify

How to Simplify

Sign

More Examples

Exercise 1

The(r+1)th term

When to Use the formula

Using the formula

Determining Value of r

More Examples

Exercise 2

Ascending/Descending Powers

Formula for (1+x)^n

Expand until

Examples

Multiplying with Polynomial/Another Expansion

Expansion of (a+b+c)^n

Exercise 3

The(r+1)th term (Part 2)

Exercise 4

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