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

Explanation

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

\[ (21C_1 - 10C_1) + (21C_2 - 10C_2) + (21C_3 - 10C_3) + \ldots + (21C_{10} - 10C_{10}) \]

### Step 1: Rewrite the Expression
We can rewrite the expression as:

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

### Step 2: Factor Out the Summation
This can be expressed as:

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

### Step 3: Use the Binomial Theorem
According to the binomial theorem, we know that:

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

Thus, for \(n=21\):

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

And for \(n=10\):

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

### Step 4: Evaluate the Sums
Now, we need to find:

\[ \sum_{k=1}^{10} 21C_k = \sum_{k=0}^{21} 21C_k - 21C_0 - \sum_{k=11}^{21} 21C_k \]

From the binomial theorem, we have:

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

So:

\[ \sum_{k=1}^{10} 21C_k = 2^{21} - 21C_0 - \sum_{k=11}^{21} 21C_k \]

### Step 5: Calculate \(21C_0\) and the Remaining Terms
We know \(21C_0 = 1\). The remaining terms can be paired:

\[ 21C_k + 21C_{21-k} = 2 \cdot 21C_k \text{ for } k=1 \text{ to } 10 \]

Thus:

\[ \sum_{k=11}^{21} 21C_k = \sum_{k=0}^{10} 21C_k = 2^{21} - 1 \]

### Step 6: Substitute Back
Now substituting back into our expression, we have:

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

### Step 7: Calculate \(10C_k\)
Now we compute:

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

### Step 8: Final Calculation
Putting it all together:

\[ \sum_{k=1}^{10} (21C_k - 10C_k) = (2^{20}) - (2^{10} - 1) = 2^{20} - 2^{10} + 1 \]

Thus, the final answer is:

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