Chapter 12.5, Problem 44AYU

Use the Binomial Theorem to find the numerical value of ${\left(0.998\right)}^6$ correct to five decimal places.

To determine

To find: The numerical value of ${\left(0.998\right)}^6$ correct to five decimal places.

Answer to Problem 44AYU

Solution:

The numerical value of ${\left(0.998\right)}^6$ using Binomial Theorem is $0.98806$

Explanation of Solution

Given: Expression is given as ${\left(0.998\right)}^6$.

It can be written as ${\left(1 - 0.002\right)}^5 = {\left(1 - 2 \times {10}^{- 3}\right)}^6$

Formula used:

The Binomial Theorem:

Let $x$ and $a$ be real numbers. For any positive integer $n$, we have

$\begin{array}{l}{\left(x + a\right)}^n = \left(\begin{array}{c}n\\ 0\end{array}\right)x^n + \left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n - 1} + \cdots + \left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n - j} + \cdots + \left(\begin{array}{c}n\\ n\end{array}\right)a^n\\ {\left(x + a\right)}^n = {\displaystyle \sum _{k = 0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n - j}{a}^j}\end{array}$

Here $x = 1,a = {10}^{- 3}$ and $n = 5$. Applying Binomial theorem,

$\begin{array}{l}{\left(1 - 2 \times {10}^{- 3}\right)}^6 = \left(\begin{array}{c}6\\ 0\end{array}\right){\left(1\right)}^6 + \left(\begin{array}{c}6\\ 1\end{array}\right)\left(- 2 \times {10}^{- 3}\right){\left(1\right)}^5 + \left(\begin{array}{c}6\\ 2\end{array}\right){\left(- 2 \times {10}^{- 3}\right)}^2{\left(1\right)}^4\\ + \left(\begin{array}{c}6\\ 3\end{array}\right){\left(- 2 \times {10}^{- 3}\right)}^3{\left(1\right)}^3 + \left(\begin{array}{c}6\\ 4\end{array}\right){\left(- 2 \times {10}^{- 3}\right)}^4{\left(1\right)}^2 + \left(\begin{array}{c}6\\ 5\end{array}\right){\left(- 2 \times {10}^{- 3}\right)}^5\left(1\right) + \left(\begin{array}{c}6\\ 6\end{array}\right){\left(- 2 \times {10}^{- 3}\right)}^6\end{array}$

$\begin{array}{l}{\left(1 - 2 \times {10}^{- 3}\right)}^6 = 1.1 + 6. - 2.\left({10}^{- 3}\right).1 + \mathrm{15.4.}\left({10}^{- 6}\right).1 + 20. - 8.\left({10}^{- 9}\right).1\\ + \mathrm{15.16.}\left({10}^{- 12}\right).1 + 6. - 32.\left({10}^{- 15}\right) + \mathrm{1.64.}\left({10}^{- 18}\right)\end{array}$

$\begin{array}{l}{\left(1 - 2 \times {10}^{- 3}\right)}^6 = 1 - \frac{12}{{10}^3} + \frac{60}{{10}^6} - \frac{160}{{10}^9} + \frac{240}{{10}^{12}} - \frac{192}{{10}^{15}} + \frac{64}{{10}^{18}}\\ = 1 - 0.012 + 0.000064 - \frac{160}{{10}^9} + \frac{240}{{10}^{12}} - \frac{192}{{10}^{15}} + \frac{64}{{10}^{18}}\end{array}$

Adding first three terms will give 6 decimals. Therefore, correcting the solution to five decimal point becomes,

${\left(1 - 2 \times {10}^{- 3}\right)}^6\cong 1 - 0.012 + 0.000064 = 0.98806$