91Ó°ÊÓ

(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 ${(1-2×{10}^{-3})}^6$

now we can apply Binomial Theorem to expand the given expression

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

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

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

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