Chapter 8.4, Problem 105E

To determine

To approximate: the quantity accurate to three decimal places.

Answer to Problem 105E

  $510,568.785$

Explanation of Solution

Given information:

Given quantity

  ${\left(2.99\right)}^{12}$

Calculation:

From binomial theorem expansion,

  ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y{\mathrm{...}}_n{C}_r{x}^{n-r}{y}^r\mathrm{...}+nx{y}^{n-1}+y^n$

Consider ${\left(2.99\right)}^{12}$ . The quantity ${\left(2.99\right)}^{12}$ can be written as ${\left(3-0.01\right)}^{12}$ . The value of ${\left(2.99\right)}^{12}$ is given by,

  $\begin{array}{}$ {\left(2.99\right)}^{12}=3^{12}-12{(3)}^{11}(0.01)+66{(3)}^{10}{(0.01)}^2-220{(3)}^9{(0.01)}^3+495{(3)}^8{(0.01)}^4-792{(3)}^7{(0.01)}^5$$ +924{(3)}^6{(0.01)}^6-792{(3)}^5{(0.01)}^7+495{(3)}^4{(0.01)}^8-220$$ {(3)}^3{(0.01)}^9+66{(3)}^2{(0.01)}^{10}-12{(3)}^1{(0.01)}^{11}+{(0.01)}^{12}$\end{array}$

Neglect the higher order terms as we are concerned with three decimal places only. Therefore, the value of ${\left(2.99\right)}^{12}$ is given by,

  $\begin{array}{}$ {\left(2.99\right)}^{12}=3^{12}-12{(3)}^{11}(0.01)+66{(3)}^{10}{(0.01)}^2-220{(3)}^9{(0.01)}^3+495{(3)}^8{(0.01)}^4-792{(3)}^7{(0.01)}^5$$ +924{(3)}^6{(0.01)}^6-792{(3)}^5{(0.01)}^7+495{(3)}^4{(0.01)}^8-220$$ {(3)}^3{(0.01)}^9+66{(3)}^2{(0.01)}^{10}-12{(3)}^1{(0.01)}^{11}+{(0.01)}^{12}$$ \approx 531,441-21,257.64+389.7234-4.3303+0.0325-0.0002+\mathrm{.....}$$ \approx 510,568.785$\end{array}$