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
$= \frac{1}{81^{n}}[^{2n}C_{0}-\,{}^{2n}C_{1}\,10^{1}+\,{}^{2n}C_{2}\,10^{2}$
$-\,{}^{2n}C_{3}\,10^{3}+.... +\,{}^{2n}C_{2n}\,10^{2n}]$
$= \frac{1}{81^{n}}\left[1-10\right]^{2n}= \frac{\left(-9\right)^{2n}}{81^{n}} = \frac{81^{n}}{81^{n}}=1$