Question

The value of $\left(\frac{1}{{81}^{\mathrm{n}}}\right)-\left(\frac{10}{{81}^{\mathrm{n}}}\right)·{}^{2\mathrm{n}}{\mathrm{C}}_1+\left(\frac{{10}^2}{{81}^{\mathrm{n}}}\right)·{}^{2\mathrm{n}}{\mathrm{C}}_2-\left(\frac{{10}^3}{{81}^{\mathrm{n}}}\right)·{}^{2\mathrm{n}}{\mathrm{C}}_3+\left(\frac{{10}^{2\mathrm{n}}}{{81}^{\mathrm{n}}}\right)$is

• A. $2$
• B. $0$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D

$1$

The explanation for the correct answer.

Given: $\left(\frac{1}{{81}^{\mathrm{n}}}\right)-\left(\frac{10}{{81}^{\mathrm{n}}}\right)·{}^{2\mathrm{n}}{\mathrm{C}}_1+\left(\frac{{10}^2}{{81}^{\mathrm{n}}}\right)·{}^{2\mathrm{n}}{\mathrm{C}}_2-\left(\frac{{10}^3}{{81}^{\mathrm{n}}}\right)·{}^{2\mathrm{n}}{\mathrm{C}}_3+......+\left(\frac{{10}^{2\mathrm{n}}}{{81}^{\mathrm{n}}}\right)$

Take $\frac{1}{{81}^{\mathrm{n}}}$ from the above expression

$=\frac{1}{{81}^{\mathrm{n}}}\left[1-10·{}^{2\mathrm{n}}{\mathrm{C}}_1+{10}^2·{}^{2\mathrm{n}}{\mathrm{C}}_2-{10}^3·{}^{2\mathrm{n}}{\mathrm{C}}_3+....+{10}^{2\mathrm{n}}\right]$

Binomial expansion of ${\left(1+\mathrm{x}\right)}^{2\mathrm{n}}={}^{2\mathrm{n}}{\mathrm{C}}_0+\mathrm{x}·{}^{2\mathrm{n}}{\mathrm{C}}_1+{\mathrm{x}}^2·{}^{2\mathrm{n}}{\mathrm{C}}_2+.....+{\mathrm{x}}^{2\mathrm{n}}$

$$\begin{gathered} \frac{1}{{81}^{\mathrm{n}}}{\left(1-10\right)}^{2\mathrm{n}}\left[\mathrm{Using}\mathrm{above}\mathrm{expansion}\right] \\ =\frac{9^{2\mathrm{n}}}{{81}^{\mathrm{n}}} \\ =\frac{{81}^{\mathrm{n}}}{{81}^{\mathrm{n}}} \\ =1 \end{gathered}$$

Hence $o\mathrm{ption}\left(\mathrm{D}\right)$ is the correct answer.