Chapter 12.5, Problem 43AYU

Use the Binomial Theorem to find the numerical value of ${\left(1.001\right)}^5$ correct to five decimal places.

[Hint: ${\left(1.001\right)}^5={\left(1+{10}^{-3}\right)}^5$]

To determine

To find: The numerical value of ${\left(1.001\right)}^5$ correct to five decimal places.

Answer to Problem 43AYU

Solution:

The numerical value of ${\left(1.001\right)}^5$ using Binomial Theorem is $1.00501$

Explanation of Solution

Given:

Expression is given as ${\left(1.001\right)}^5$.

It can be written as ${\left(1 + 0.001\right)}^5 = {\left(1 + {10}^{- 3}\right)}^5$

Formula used:

The Binomial Theorem:

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

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

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

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

$\begin{array}{l}{\left(1 + {10}^{- 3}\right)}^5 = 1.1 + 5.\left({10}^{- 3}\right).1 + 10{\left({10}^{- 3}\right)}^{2.1} + 10{\left({10}^{- 3}\right)}^{3.1} + 5{\left({10}^{- 3}\right)}^{4.1} + 1.{\left({10}^{- 3}\right)}^5\\ {\left(1 + {10}^{- 3}\right)}^5 = 1 + \frac{5}{{10}^3} + \frac{10}{{10}^6} + \frac{10}{{10}^9} + \frac{5}{{10}^{12}} + \frac{1}{{10}^{15}}\end{array}$

${\left(1 + {10}^{- 3}\right)}^5 = 1 + \frac{5}{{10}^3} + \frac{1}{{10}^5} + \frac{1}{{10}^8} + \frac{5}{{10}^{12}} + \frac{1}{{10}^{15}}$

${\left(1 + {10}^{- 3}\right)}^5 = 1.005010010005001$

Correcting the solution to five decimal point,

${\left(1 + {10}^{- 3}\right)}^5\cong 1.00501$