Use the Binomial Theorem to find the numerical value of

(0.998)⁶ correct to five decimal places.

(0.998)⁶ is given.

We have to find the numerical value of (0.998)⁶ correct to five decimal places.

we can rewrite the given expression as ${(1-2\times {10}^{-3})}^6$

now we can apply Binomial Theorem to expand the given expression

$$\begin{gathered} Letxandaberealnumbers.Foranypositiveintegern,wehave \\ {(x+a)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\left(\begin{array}{c}n\\ 2\end{array}\right)a^2{x}^{n-2}+.......+\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+.....+\left(\begin{array}{c}n\\ n\end{array}\right)a^n \\ {(1-2\times {10}^{-3})}^6=\left(\begin{array}{c}6\\ 0\end{array}\right){{(-2\times {10}^{-3})}^0\left(1\right)}^6+\left(\begin{array}{c}6\\ 1\end{array}\right){{(-2\times {10}^{-3})}^1\left(1\right)}^5+\left(\begin{array}{c}6\\ 2\end{array}\right){{(-2\times {10}^{-3})}^2\left(1\right)}^4+\left(\begin{array}{c}6\\ 3\end{array}\right){{(-2\times {10}^{-3})}^3\left(1\right)}^3+\left(\begin{array}{c}6\\ 4\end{array}\right){{(-2\times {10}^{-3})}^4\left(1\right)}^2+\left(\begin{array}{c}6\\ 5\end{array}\right){{(-2\times {10}^{-3})}^5\left(1\right)}^1+\left(\begin{array}{c}6\\ 6\end{array}\right){{(-2\times {10}^{-3})}^6\left(1\right)}^0 \\ =\left(\begin{array}{c}6\\ 0\end{array}\right)-\left(\begin{array}{c}6\\ 1\end{array}\right)(2\times {10}^{-3})+\left(\begin{array}{c}6\\ 2\end{array}\right){(2\times {10}^{-3})}^2-\left(\begin{array}{c}6\\ 3\end{array}\right){(2\times {10}^{-3})}^3+\left(\begin{array}{c}6\\ 4\end{array}\right){(2\times {10}^{-3})}^4-\left(\begin{array}{c}6\\ 5\end{array}\right){(2\times {10}^{-3})}^5+\left(\begin{array}{c}6\\ 6\end{array}\right){(2\times {10}^{-3})}^6 \end{gathered}$$

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

$$\begin{gathered} {\left((1-2\times {10}^{-3}\right)}^6=\left(\begin{array}{c}6\\ 0\end{array}\right)-\left(\begin{array}{c}6\\ 1\end{array}\right)(2\times {10}^{-3})+\left(\begin{array}{c}6\\ 2\end{array}\right){(2\times {10}^{-3})}^2-\left(\begin{array}{c}6\\ 3\end{array}\right){(2\times {10}^{-3})}^3+\left(\begin{array}{c}6\\ 4\end{array}\right){(2\times {10}^{-3})}^4-\left(\begin{array}{c}6\\ 5\end{array}\right){(2\times {10}^{-3})}^5+\left(\begin{array}{c}6\\ 6\end{array}\right){(2\times {10}^{-3})}^6 \\ numericalvalueof{(1-2\times {10}^{-3})}^6=\left(\begin{array}{c}6\\ 0\end{array}\right)-\left(\begin{array}{c}6\\ 1\end{array}\right)(2\times {10}^{-3})+\left(\begin{array}{c}6\\ 2\end{array}\right){(2\times {10}^{-3})}^2-\left(\begin{array}{c}6\\ 3\end{array}\right){(2\times {10}^{-3})}^3+\left(\begin{array}{c}6\\ 4\end{array}\right){(2\times {10}^{-3})}^4-\left(\begin{array}{c}6\\ 5\end{array}\right){(2\times {10}^{-3})}^5+\left(\begin{array}{c}6\\ 6\end{array}\right){(2\times {10}^{-3})}^6 \\ =1-\frac{6!}{5!1!}\times 2\times {10}^{-3}+\frac{6!}{4!2!}\times 4\times {10}^{-6}-\frac{6!}{3!3!}\times 8\times {10}^{-9}+\frac{6!}{2!4!}\times 16\times {10}^{-12}-\frac{6!}{1!5!}\times 32\times {10}^{-15}+64\times {10}^{-18} \\ =1-\frac{6\times 5!}{1\times 5!}\times {10}^{-3}+\frac{6\times 5\times 4!}{1\times 2\times 4!}\times 4\times {10}^{-6}-\frac{6\times 5\times 4\times 3!}{1\times 2\times \times 3\times 3!}\times 8\times {10}^{-9}+\frac{6\times 5\times 4!}{1\times 2\times 4!}\times 16\times {10}^{-12}-\frac{6\times 5!}{1\times 5!}\times 32\times {10}^{-15}+64\times {10}^{-18} \\ =1-6\times {10}^{-3}+60\times {10}^{-6}-160\times {10}^{-9}+240\times {10}^{-12}-192\times {10}^{-15}+64\times {10}^{-18} \\ {(1-2\times {10}^{-3})}^6=1-0.006+0.00006+-160\times {10}^{-9}+240\times {10}^{-12}-192\times {10}^{-15}+64\times {10}^{-18} \\ Numericalvaluescorrecttothe5decimalplacesis0.99406 \end{gathered}$$

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