# How do you use the binomial theorem to approximate (2.99)^12?

Sep 4, 2017

${2.99}^{12} \approx 510568.79$

#### Explanation:

Note that:

${\left(2.99\right)}^{12} = {\left(3 - 0.01\right)}^{12}$

is of the form ${\left(a + b\right)}^{n}$, with $a = 3$, $b = 0.01$ and $n = 12$.

The binomial theorem tells us that:

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

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

Instead of calculating these binomial coefficients directly, we can pick them out from the row of Pascal's triangle that starts $1 , 12 ,$... That is:

$1 , 12 , 66 , 220 , 495 , 792 , 924 , 792 , 495 , 220 , 66 , 12 , 1$

Hence:

${2.99}^{12} = {\left(3 - 0.01\right)}^{12}$

$\textcolor{w h i t e}{{2.99}^{12}} \approx {3}^{12} - 12 \left({3}^{11}\right) \left(0.01\right) + 66 \left({3}^{10}\right) {\left(0.01\right)}^{2} - 220 \left({3}^{9}\right) {\left(0.01\right)}^{3} + 495 \left({3}^{8}\right) {\left(0.01\right)}^{4}$

$\textcolor{w h i t e}{{2.99}^{12}} = 531441 - 0.12 \left(177147\right) + 0.0066 \left(59049\right) - 0.000220 \left(19683\right) + 0.00000495 \left(6561\right)$

$\textcolor{w h i t e}{{2.99}^{12}} = 531441 - 21257.64 + 389.7234 - 4.330260 + 0.03247695$

$\textcolor{w h i t e}{{2.99}^{12}} \approx 510568.79$