Find the value of `(.^(10)C_(10))+(.^(10)C_(0)+.^(10)C_(1))+(.^(10)C_(0)+.^(10)C_(1)+.^(10)C_(2))+"...."+(.^(10)C_(0)+.^(10)C_(1)+.^(10)C_(2)+"....." + .^(10)C_(9))`.

Explanation

To solve the problem, we need to evaluate the expression given in the question step by step.

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression can be rewritten as:
\[ \sum_{k=0}^{10} \binom{10}{k} \cdot (10-k) \]
This represents the sum of combinations where we are summing the number of ways to choose `k` items from `10`, multiplied by the number of items left, which is `10-k`.

2. **Rearranging the Terms**:
We can rearrange the expression to make it clearer:
\[ \sum_{k=0}^{10} (10-k) \binom{10}{k} = 10 \sum_{k=0}^{10} \binom{10}{k} - \sum_{k=0}^{10} k \binom{10}{k} \]

3. **Using the Binomial Theorem**:
According to the Binomial Theorem, we know that:
\[ \sum_{k=0}^{n} \binom{n}{k} = 2^n \]
For \( n = 10 \):
\[ \sum_{k=0}^{10} \binom{10}{k} = 2^{10} = 1024 \]

4. **Calculating the Second Sum**:
The second sum, \( \sum_{k=0}^{10} k \binom{10}{k} \), can be simplified using the identity:
\[ k \binom{n}{k} = n \binom{n-1}{k-1} \]
Thus,
\[ \sum_{k=0}^{10} k \binom{10}{k} = 10 \sum_{k=1}^{10} \binom{9}{k-1} = 10 \cdot 2^9 = 10 \cdot 512 = 5120 \]

5. **Substituting Back**:
Now substituting back into our rearranged expression:
\[ 10 \cdot 1024 - 5120 = 10240 - 5120 = 5120 \]

6. **Final Result**:
Therefore, the value of the original expression is:
\[ \boxed{5120} \]