The value of `(.^(21)C_(1) - .^(10)C_(1)) + (.^(21)C_(2) - .^(10)C_(2)) + (.^(21)C_(3) - .^(10)C_(3)) + (.^(21)C_(4) - .^(10)C_(4)) + … + (.^(21)C_(10) - .^(10)C_(10))` is

Options

A. `2^(20) - 2^(10)`
B. `2^(21) - 2^(11)`
C. `2^(21) - 2^(10)`
D. `2^(20) - 2^(9)`

Explanation

To solve the given problem, we need to evaluate the expression:

\[ \sum_{k=1}^{10} \left( \binom{21}{k} - \binom{10}{k} \right) \]

### Step 1: Break down the summation
We can express the summation as:

\[ \sum_{k=1}^{10} \binom{21}{k} - \sum_{k=1}^{10} \binom{10}{k} \]

### Step 2: Evaluate the first summation
The first summation, \(\sum_{k=1}^{10} \binom{21}{k}\), represents the sum of the first 10 binomial coefficients from the expansion of \((1 + 1)^{21}\). By the binomial theorem, we know that:

\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]

Thus,

\[ \sum_{k=0}^{21} \binom{21}{k} = 2^{21} \]

However, we want only the first 10 terms. To find this, we can use the identity:

\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]

This means:

\[ \sum_{k=0}^{21} \binom{21}{k} = 2^{21} \]

But we need to subtract the terms from \(k=11\) to \(k=21\):

\[ \sum_{k=11}^{21} \binom{21}{k} = \sum_{k=0}^{21} \binom{21}{k} - \sum_{k=0}^{10} \binom{21}{k} \]

Using the symmetry property of binomial coefficients, we know:

\[ \sum_{k=11}^{21} \binom{21}{k} = \sum_{k=0}^{10} \binom{21}{k} \]

Thus, we can conclude:

\[ \sum_{k=0}^{10} \binom{21}{k} = \frac{1}{2} \cdot 2^{21} = 2^{20} \]

So,

\[ \sum_{k=1}^{10} \binom{21}{k} = 2^{20} - 1 \]

### Step 3: Evaluate the second summation
Now we evaluate the second summation, \(\sum_{k=1}^{10} \binom{10}{k}\):

\[ \sum_{k=0}^{10} \binom{10}{k} = 2^{10} \]

Thus,

\[ \sum_{k=1}^{10} \binom{10}{k} = 2^{10} - 1 \]

### Step 4: Combine the results
Now we can combine the results from both summations:

\[ \sum_{k=1}^{10} \left( \binom{21}{k} - \binom{10}{k} \right) = (2^{20} - 1) - (2^{10} - 1) \]

This simplifies to:

\[ 2^{20} - 2^{10} \]

### Final Result
Thus, the final value of the expression is:

\[ 2^{20} - 2^{10} \]