The value of `(1)/(81^(n))-(10)/(81^(n)).""^(2n)C_(2)-(10^(3))/(81^(n)).""^(2n)C_(3)+…+((10)^(2n))/(81^(n))` is

Options

A. 2
B. 1
C. 0
D. `1//2`

Answer

To solve the expression

\[ \frac{1}{81^n} - \frac{10}{81^n} \binom{2n}{2} + \frac{10^3}{81^n} \binom{2n}{3} - \ldots + \frac{10^{2n}}{81^n}, \]

we can recognize that this is a series that can be expressed in terms of the binomial expansion.

### Step 1: Identify the series
The series can be rewritten as:

\[ S = \sum_{k=0}^{2n} (-1)^k \frac{10^k}{81^n} \binom{2n}{k}. \]

### Step 2: Factor out the common term
Notice that \( \frac{1}{81^n} \) is a common factor in each term of the series. We can factor it out:

\[ S = \frac{1}{81^n} \sum_{k=0}^{2n} (-1)^k 10^k \binom{2n}{k}. \]

### Step 3: Recognize the binomial theorem
The sum inside the parentheses is a binomial expansion. According to the binomial theorem:

\[ \sum_{k=0}^{m} (-1)^k \binom{m}{k} x^k = (1 - x)^m. \]

In our case, \( m = 2n \) and \( x = 10 \):

\[ \sum_{k=0}^{2n} (-1)^k \binom{2n}{k} 10^k = (1 - 10)^{2n} = (-9)^{2n}. \]

### Step 4: Substitute back into the equation
Now substituting this back into our expression for \( S \):

\[ S = \frac{1}{81^n} \cdot (-9)^{2n}. \]

### Step 5: Simplify the expression
We can simplify \( (-9)^{2n} \):

\[ (-9)^{2n} = 9^{2n}. \]

So we have:

\[ S = \frac{9^{2n}}{81^n}. \]

### Step 6: Simplify further
Since \( 81 = 9^2 \), we can rewrite \( 81^n \) as \( (9^2)^n = 9^{2n} \). Thus,

\[ S = \frac{9^{2n}}{9^{2n}} = 1. \]

### Final Answer
The value of the given expression is:

\[ \boxed{1}. \]