Binomial Theorem (DP IB Analysis & Approaches (AA): SL): Revision Note

Binomial theorem

What is the binomial theorem?

• The binomial theorem gives you the expansion of ${\left(a+b\right)}^n$ for different positive integer powers of $n$:

• where is the binomial coefficient

  • and the factorial symbol $!$ means, for example:

    • $4!=4\times 3\times 2\times 1=24$

How do I use the binomial theorem?

• An example of using the binomial theorem is to expand ${\left(a+b\right)}^5$

  • Substitute in $n=5$

  • 

  • where

    • 

    • 

    • 

    • 

  • giving

    • ${\left(a+b\right)}^5=a^5+5a^{4}b+10a^3{b}^2+10a^2{b}^3+5a{b}^4+b^5$

• The total of the powers of each term must equal the power of the binomial, $n$

  • e.g. the total of the power of $10a^3{b}^2$ is $3+2=5$

• The binomial theorem saves you from having to expand by hand

  • ${\left(a+b\right)}^5=\left(a+b\right)\left(a+b\right)\left(a+b\right)\left(a+b\right)\left(a+b\right)$

How do I expand harder binomials?

• To expand harder binomials like ${\left(1+2x\right)}^5$, put brackets around the terms

  • then put the powers outside the brackets

    • e.g. ${\left(2x\right)}^3$

  • and use index laws

    • e.g. ${\left(2x\right)}^3=8x^3$

• To expand negative terms like ${\left(1-3y\right)}^n$, use brackets

  • e.g. ${\left(-3y\right)}^2=9y^2$

How do I find a particular term in a binomial expansion?

• To find a particular term

  • either expand the whole binomial

  • or use the general term formula $\begin{array}{rcl}& & {}_{ }{}^n{\mathrm{C}}_r{a}^{n-r}{b}^r\end{array}$

    • and form an equation to find $r$

• For example, find the term in $x^{14}$ in the expansion of ${\left(2x+x^3\right)}^8$

  • first substitute $n=8$, $a=\left(2x\right)$ and $b=\left(x^3\right)$ into $\begin{array}{rcl}& & {}_{ }{}^n{\mathrm{C}}_r{a}^{n-r}{b}^r\end{array}$

    • $\begin{array}{rcl}& & {}_{ }{}^8{\mathrm{C}}_r{\left(2x\right)}^{8-r}{\left(x^3\right)}^r\end{array}$

  • then just look at the power of $x$ only

    • $x^{8-r}\times x^{3r}=x^{8-r+3r}=x^{8+2r}$

  • For this to equal $x^{14}$, you need $8+2r=14$

    • which solves to give $r=3$

  • Substitute $r=3$ back into $\begin{array}{rcl}& & {}_{ }{}^8{\mathrm{C}}_r{\left(2x\right)}^{8-r}{\left(x^3\right)}^r\end{array}$ to get the whole term

    • $\begin{array}{rcl}{}_{ }{}^8{\mathrm{C}}_3{\left(2x\right)}^5{\left(x^3\right)}^3& =& {}_{ }{}^8{\mathrm{C}}_3{2}^5{x}^5{x}^9={}_{ }{}^8{\mathrm{C}}_{3}32x^{14}\end{array}$

    • and

    • giving $\begin{array}{rcl}& & 56\times 32x^{14}\end{array}$

  • The answer is $1792x^{14}$