# How do you use the binomial theorem to approximate (1.02)^8?

May 12, 2017

See below.

#### Explanation:

Original question: Use binomial theorem to find ${\left(1.02\right)}^{8}$

Consider the binomial theorem:
${\left(a + b\right)}^{n} = \left(\begin{matrix}n \\ 0\end{matrix}\right) {a}^{n} {b}^{0} + \left(\begin{matrix}n \\ 1\end{matrix}\right) {a}^{n - 1} {b}^{1} + \ldots + \left(\begin{matrix}n \\ n - 1\end{matrix}\right) {a}^{1} {b}^{n} - 1 + \left(\begin{matrix}n \\ n\end{matrix}\right) {a}^{0} {b}^{n}$
where $n$ is a positive integer and $\left(\begin{matrix}x \\ y\end{matrix}\right)$ is $x$ choose $y$.

Since the binomial theorem only works on values in the form of a binomial: Consider that $1.02 = 1 + 0.02 = 1 + \frac{1}{50}$

So, by substituting $1.02 = 1 + \frac{1}{50}$, we get:
${\left(1 + \frac{1}{50}\right)}^{8}$
By applying the binomial theorem, we get:
$= \left(\begin{matrix}8 \\ 0\end{matrix}\right) + \left(\begin{matrix}8 \\ 1\end{matrix}\right) \left(\frac{1}{50}\right) + \left(\begin{matrix}8 \\ 2\end{matrix}\right) {\left(\frac{1}{50}\right)}^{2} + \left(\begin{matrix}8 \\ 3\end{matrix}\right) {\left(\frac{1}{50}\right)}^{3} + \left(\begin{matrix}8 \\ 4\end{matrix}\right) {\left(\frac{1}{50}\right)}^{4} + \left(\begin{matrix}8 \\ 5\end{matrix}\right) {\left(\frac{1}{50}\right)}^{5} + \left(\begin{matrix}8 \\ 6\end{matrix}\right) {\left(\frac{1}{50}\right)}^{6} + \left(\begin{matrix}8 \\ 7\end{matrix}\right) {\left(\frac{1}{50}\right)}^{7} + \left(\begin{matrix}8 \\ 8\end{matrix}\right) {\left(\frac{1}{50}\right)}^{8}$
$= 1 + \frac{8}{50} + \frac{28}{50} ^ 2 + \frac{56}{50} ^ 3 + \frac{70}{50} ^ 4 + \frac{56}{50} ^ 5 + \frac{28}{50} ^ 6 + \frac{8}{50} ^ 7 + \frac{1}{50} ^ 8$
$= \frac{45767944570401}{39062600000000} \approx 1.17166$ rounded to 5 decimal places

We can confirm this result:
${\left(1.02\right)}^{8} \approx 1.17166$ rounded to 5 decimal places