# Binomial Expansion – negative & fractional powers

This page details the more advanced use of binomial expansion. You should be familiar with all of the material from the more basic Binomial Expansion page first.

Recall that the first formula provided in the Edexcel formula booklet is:

$(a+b)^n=a^n+\left(\begin{array}{c}n\\1\end{array}\right)a^{n-1}b+\left(\begin{array}{c}n\\2\end{array}\right)a^{n-2}b^2+…+\left(\begin{array}{c}n\\r\end{array}\right)a^{n-r}b^r+…+b^n, \hspace{20pt}\left(n\in{\mathbb N}\right)$

where $\left(\begin{array}{c} n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}$. However, this formula is only valid for positive integer $n$.

In addition to this, the booklet also provides a second formula for negative and fractional powers:

$\left(1+x\right)^n=1+nx+\frac{n(n-1)}{1\times 2}x^2+…+\frac{n(n-1)…(n-r+1)}{1\times 2\times …\times r}x^r+…,\hspace{20pt}\left(\vert x\vert <1, n\in {\mathbb R}\right)$

The first formula is only valid for positive integer $n$ but this formula is valid for all $n$. This includes negative and fractional powers. Note, however, the formula is not valid for all values of $x$. As stated, the $x$ values must be between -1 and 1.

### Range of Validity for Binomial Expansions

As stated above, the second formula for binomial expansion in the Edexcel Formula Booklet is only valid for $\vert x\vert <1$. This is because, unlike for positive integer $n$, these expansions have an infinite number of terms (as indicted by the … in the formula). Subsequently, we require the series to converge as the powers of $x$ become large. For this to happen, we must have $\vert x\vert <1$. Questions may ask you to find the binomial expansion without explicitly stating the value of $n$ and ask you to identify the values for which the expansion is valid. See the Identifying the Power Example.

Also notice that in this second formula there is a very specific format inside the brackets – it must be 1 plus something. Therefore, if there is something other than 1 inside these brackets, the coefficient must be factored out. Do this by first writing $(a+bx)^n=\left(a\left(1+\frac{bx}{a}\right)\right)^n=a^n\left(1+\frac{bx}{a}\right)^n$. Then find the expansion of $\left(1+\frac{bx}{a}\right)^n$ using the formula. Do this by replacing all $x$ with $\frac{bx}{a}$. This inevitably changes the range of validity. It follows that this expansion will be valid for $\left\vert \frac{bx}{a}\right\vert <1$ or $\vert x\vert <\frac{a}{b}$. See the Factoring Out Example.

At more advanced levels, questions may ask you to use partial fractions first. See the Using Partial Fractions question.

## Binomial Expansion Examples

### Identifying the Power

- Find the first four terms in ascending powers of $x$ of the binomial expansion of $\frac{1}{(1+2x)^2}$.
- State the range of validity for your expansion.

- Firstly, write the expression as $\left(1+2x\right)^{-2}$. The power $n=-2$ is negative and so we must use the second formula. We can then find the expansion by setting $n=-2$ and replacing all $x$ with $2x$: $\begin{array}{l}&&\left(1+2x\right)^{-2}\\&=&1-2(2x)+\frac{-2(-3)}{1\times 2}(2x)^2+\frac{-2(-3)(-4)}{1\times 2\times 3}(2x)^3+…\\&=&1-4x+12x^2-32x^3+…\end{array}$
- We find the range of validity by replacing $x$ with $2x$ in the expression $\vert x\vert <1$ to give $\vert 2x\vert <1$. It follows that the expansion is valid for $\vert x\vert <\frac{1}{2}$. We can also write this as $-\frac{1}{2}<x<\frac{1}{2}$.

### Factoring out

Find the first three terms, in ascending powers of $x$, of the expansion of $\sqrt{4-3x}$. State the range of validity of your expansion and use it to find an approximation to $\sqrt{3.7}$.

Begin by writing the expression as $\left(4-3x\right)^{\frac{1}{2}}$. The power $n=\frac{1}{2}$ is fractional so we must use the second formula. However, this first requires us to remove a factor of 4:

$\left(4-3x\right)^{\frac{1}{2}}=\left(4\left(1-\frac{3x}{4}\right)\right)^{\frac{1}{2}}=4^{\frac{1}{2}}\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}$

Note that the expansion of $\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}$ is given by:

$\begin{array}{l}\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}&=&1+\frac{1}{2}\left(-\frac{3x}{4}\right)+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{1\times 2}\left(-\frac{3x}{4}\right)^2+…\\&=&1-\frac{3}{8}x-\frac{9}{128}x^2+…\end{array}$

Hence, multiplying by the factor of $4^{\frac{1}{2}}=2$ gives:

$\left(4-3x\right)^{\frac{1}{2}}=2\left(1-\frac{3x}{4}\right)^{\frac{1}{2}}\approx 2-\frac{3}{4}x-\frac{9}{64}x^2$

This expansion is valid for $\vert -\frac{3x}{4}\vert <1$, that is $\vert x\vert <\frac{4}{3}$.

Finally, by setting $x=0.1$, we can find an approximation to $\sqrt{3.7}$:

$\left(3.7\right)^{\frac{1}{2}}\approx 2-\frac{3}{4}\times 0.1-\frac{9}{64}\times 0.1^2\approx 1.9246 $

to 4 decimal places. Note that we can do this since $-\frac{4}{3}<0.1<\frac{4}{3}$. The exact value of $\sqrt{3.7}=1.9235$ to 4 decimal places, which is a reasonable approximation.

### Using Partial Fractions

- Express $f(x)=\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}$ as partial fractions.
- Show that the quadratic approximation to $f(x)$ is given by $f(x)\approx 3+\frac{13}{2}x+\frac{19}{4}x^2$.
- State the range of values of $x$ for which this approximation is valid.

- Firstly, revise partial fractions by clicking here. Note that $\frac{A}{1-x}+\frac{B}{1+\frac{1}{2}x}=\frac{A\left(1+\frac{1}{2}x\right)+B(1-x)}{(1-x)(1+\frac{1}{2}x)}$. Hence, $A+B=3$ and $\frac{1}{2}A-B=5$. It follows that $A=\frac{16}{3}$ and $B=-\frac{7}{3}$ and so $\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}=\frac{16}{3(1-x)}-\frac{7}{3\left(1+\frac{1}{2}x\right)}$.
- From part 1, we can write $\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}=\frac{16}{3}(1-x)^{-1}-\frac{7}{3}\left(1+\frac{1}{2}x\right)^{-1}$. Now, take binomial expansions for each term: $(1-x)^{-1}=1+x+x^2+…$ $\left(1+\frac{1}{2}x\right)^{-1}=1-\frac{1}{2}x+\frac{1}{4}x^2+…$ Hence, $\begin{array}{l}\frac{3+5x}{(1-x)(1+\frac{1}{2}x)}&\approx&\frac{16}{3}(1+x+x^2)-\frac{7}{3}\left(1-\frac{1}{2}x+\frac{1}{4}x^2\right)\\&=&3+\frac{13}{2}x+\frac{19}{4}x^2\end{array}$ as given.
- The first expansion is valid for $\vert -x\vert <1$ (or $-1<x<1$) while the second expansion is valid for $\vert \frac{1}{2}x\vert <1$ (or $-2<x<2$. In order for both of these inequalities to be satisfied, we must have $-1<x<1$.