Question

Use the Binomial Theorem to find the numerical value of

(1.001)⁵ correct to five decimal places.

Short answer

The numerical value of

(1.001)⁵ correct to five decimal places is$1.00501$

Step-by-step solution

Step1. Given information 

Here(1.001)⁵ is given.

We need to find out the numerical value of (1.001)⁵ correct to five decimal places .

Step 2. Description of expanding (1.001)5

we can rewrite the given expression as (1+10^-3)⁵

now we can apply Binomial Theorem.

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.......+\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \\ {(1+{10}^{-3})}^5=\left(\begin{array}{c}5\\ 0\end{array}\right){{\left({10}^{-3}\right)}^0\left(1\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){{\left({10}^{-3}\right)}^1\left(1\right)}^4+\left(\begin{array}{c}5\\ 2\end{array}\right){{\left({10}^{-3}\right)}^2\left(1\right)}^3+\left(\begin{array}{c}5\\ 3\end{array}\right){{\left({10}^{-3}\right)}^3\left(1\right)}^2+\left(\begin{array}{c}5\\ 4\end{array}\right){{\left({10}^{-3}\right)}^4\left(1\right)}^1+\left(\begin{array}{c}5\\ 5\end{array}\right){{\left({10}^{-3}\right)}^5\left(1\right)}^0 \\ =\left(\begin{array}{c}5\\ 0\end{array}\right)+\left(\begin{array}{c}5\\ 1\end{array}\right)\left({10}^{-3}\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left({10}^{-3}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left({10}^{-3}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left({10}^{-3}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left({10}^{-3}\right)}^5 \end{gathered}$$

Step3. Finding  the numerical value of(1.001)5

The expanded form of (1.001)⁵ is

$$\begin{gathered} {(1+{10}^{-3})}^5=\left(\begin{array}{c}5\\ 0\end{array}\right)+\left(\begin{array}{c}5\\ 1\end{array}\right)\left(10-3\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left({10}^{-3}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left({10}^{-3}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left({10}^{-3}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left({10}^{-3}\right)}^5 \\ numericalvalueof{(1+{10}^{-3})}^5=\left(\begin{array}{c}5\\ 0\end{array}\right)+\left(\begin{array}{c}5\\ 1\end{array}\right)\left(10-3\right)+\left(\begin{array}{c}5\\ 2\end{array}\right){\left({10}^{-3}\right)}^2+\left(\begin{array}{c}5\\ 3\end{array}\right){\left({10}^{-3}\right)}^3+\left(\begin{array}{c}5\\ 4\end{array}\right){\left({10}^{-3}\right)}^4+\left(\begin{array}{c}5\\ 5\end{array}\right){\left({10}^{-3}\right)}^5 \\ =1+\frac{5!}{4!1!}\times {10}^{-3}+\frac{5!}{3!2!}\times {10}^{-6}+\frac{5!}{2!3!}\times {10}^{-9}+\frac{5!}{1!4!}\times {10}^{-12}+1\times {10}^{-15} \\ =1+\frac{5\times 4!}{1\times 4!}\times {10}^{-3}+\frac{5\times 4\times 3!}{1\times 2\times 3!}\times {10}^{-6}+\frac{5\times 4\times 3!}{1\times 2\times 3!}\times {10}^{-9}+\frac{5\times 4!}{1\times 4!}\times {10}^{-12}+{10}^{-15} \\ =1+5\times {10}^{-3}+10\times {10}^{-6}+10\times {10}^{-9}+5\times {10}^{-12}+{10}^{-15} \\ {(1+{10}^{-3})}^5=1+0.005+0.00001+{10}^{-8}+5\times {10}^{-12}+{10}^{-15} \\ Numericalvaluescorrecttothe5decimalplacesis1.00501. \end{gathered}$$.