# How do you use the binomial theorem to approximate (1.98)^9?

Aug 23, 2017

${1.98}^{9} \approx 467.721$

#### Explanation:

${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$

where ((n),(k)) = (n!)/((n-k)!k!)

We could use this with $a = 2$ and $b = - 0.02$, but it's probably cleaner to separate out the factor ${2}^{9} = 512$ and use $a = 1$ and $b = - 0.01$ instead.

We can find the binomial coefficients $\left(\begin{matrix}9 \\ k\end{matrix}\right)$ we need from the row of Pascal's triangle that starts $1 , 9 , 36 , \ldots$

So we find:

${1.98}^{9} = {2}^{9} \cdot {\left(1 - 0.01\right)}^{9}$

$\textcolor{w h i t e}{{1.98}^{9}} = {2}^{9} \cdot \left(1 - 9 \left(0.01\right) + 36 \left(0.0001\right) - 84 \left(0.000001\right) + \ldots\right)$

$\textcolor{w h i t e}{{1.98}^{9}} = {2}^{9} \cdot \left(1 - 0.09 + 0.0036 - 0.000084 + \ldots\right)$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{9} \cdot \left(1 - 0.09 + 0.0036 - 0.000083\right)$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{9} \cdot 0.913517$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{8} \cdot 1.827034$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{7} \cdot 3.654068$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{6} \cdot 7.308136$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{5} \cdot 14.616272$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{4} \cdot 29.232544$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{3} \cdot 58.465088$

$\textcolor{w h i t e}{{1.98}^{9}} \approx {2}^{2} \cdot 116.930176$

$\textcolor{w h i t e}{{1.98}^{9}} \approx 2 \cdot 233.860352$

$\textcolor{w h i t e}{{1.98}^{9}} \approx 467.721$