Question

The value of $\frac{1}{81^n}-\frac{10}{81^n}{}^{2n}{C}_1+\frac{10^2}{81^n}{}^{2n}{C}_2-\frac{10^3}{81^n}{}^{2n}{C}_3+.....+\frac{10^{2n}}{81^n}is$

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

Answer 1

Answer: (D)

The given expression
$=\frac{1}{81^n}{[}^{2n}{C}_0-{}^{2n}{C}_{1}10^1+{}^{2n}{C}_{2}10^2$
$-{}^{2n}{C}_{3}10^3+....+{}^{2n}{C}_{2n}10^{2n}]$
$=\frac{1}{81^n}{\left[1-10\right]}^{2n}=\frac{{\left(-9\right)}^{2n}}{81^n}=\frac{81^n}{81^n}=1$