Evaluate the expression: frac{1}{81^{n}} - {} | vedclass

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(C) The given expression is $\frac{1}{81^{n}} \left[ 1 - {}^{2n}C_1(10) + {}^{2n}C_2(10^2) - \dots + (-1)^{2n} {}^{2n}C_{2n}(10^{2n}) \right]$.Using t

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$1$

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(C) The given expression is $\frac{1}{81^{n}} \left[ 1 - {}^{2n}C_1(10) + {}^{2n}C_2(10^2) - \dots + (-1)^{2n} {}^{2n}C_{2n}(10^{2n}) \right]$.Using the binomial expansion formula $(a - b)^m = \sum_{k=0}^{m} {}^{m}C_k a^{m-k} (-b)^k$,we identify $a = 1$,$b = 10$,and $m = 2n$.The expression inside the bracket is $(1 - 10)^{2n} = (-9)^{2n}$.Substituting this back,we get $\frac{(-9)^{2n}}{81^{n}} = \frac{((-9)^2)^n}{81^n} = \frac{81^n}{81^n} = 1$.

Evaluate the expression: $\frac{1}{81^{n}} - {}^{2n}C_1 \frac{10}{81^{n}} + {}^{2n}C_2 \frac{10^2}{81^{n}} - \dots + \frac{10^{2n}}{81^{n}} = $

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