If `(1 + x)^(n) = C_(0) + C_(1) x C_(2) x^(2) +…+ C_(n) x^(n)`, then
the sum `C_(0) + (C_(0)+C_(1))+…+(C_(0) +C_(1) +…+C_(n -1))` is equal to

Options

A. `n . 2^(n)`
B. `n . 2^(n-1)`
C. `n . 2^(n-2)`
D. `n. 2^(n-3)`

Answer

To solve the problem, we need to find the sum \( S = C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + \ldots + C_{n-1}) \).

### Step-by-Step Solution:

1. **Understanding the Terms**:
The given expression can be rewritten as:
\[ S = C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + \ldots + C_{n-1}) \]
Each term in this sum is a cumulative sum of the binomial coefficients.

2. **Identifying the General Term**:
The \( r \)-th term in the sum can be expressed as:
\[ C_0 + C_1 + C_2 + \ldots + C_r \]
for \( r = 0, 1, 2, \ldots, n-1 \).

3. **Counting the Contributions**:
- \( C_0 \) appears in all \( n \) terms.
- \( C_1 \) appears in \( n-1 \) terms.
- \( C_2 \) appears in \( n-2 \) terms.
- Continuing this pattern, \( C_{n-1} \) appears in \( 1 \) term.

4. **Expressing the Sum**:
The total sum can be expressed as:
\[ S = nC_0 + (n-1)C_1 + (n-2)C_2 + \ldots + 1 \cdot C_{n-1} \]

5. **Using Binomial Coefficient Properties**:
We can express this sum in summation notation:
\[ S = \sum_{r=0}^{n-1} (n-r) C_r \]

6. **Splitting the Summation**:
This can be split into two separate sums:
\[ S = n \sum_{r=0}^{n-1} C_r - \sum_{r=0}^{n-1} r C_r \]

7. **Calculating the First Sum**:
The first sum \( \sum_{r=0}^{n-1} C_r \) is known to be:
\[ \sum_{r=0}^{n-1} C_r = 2^{n-1} \]

8. **Calculating the Second Sum**:
The second sum \( \sum_{r=0}^{n-1} r C_r \) can be calculated using the identity:
\[ \sum_{r=0}^{n} r C_r = n \cdot 2^{n-1} \]
Therefore, for \( n-1 \):
\[ \sum_{r=0}^{n-1} r C_r = (n-1) \cdot 2^{n-2} \]

9. **Substituting Back**:
Now substituting back into the expression for \( S \):
\[ S = n \cdot 2^{n-1} - (n-1) \cdot 2^{n-2} \]

10. **Simplifying**:
\[ S = n \cdot 2^{n-1} - \frac{(n-1)}{2} \cdot 2^{n-1} = \left(n - \frac{(n-1)}{2}\right) \cdot 2^{n-1} \]
\[ S = \left(\frac{2n - (n-1)}{2}\right) \cdot 2^{n-1} = \left(\frac{n + 1}{2}\right) \cdot 2^{n-1} \]

11. **Final Result**:
Therefore, the final result for the sum is:
\[ S = n \cdot 2^{n-1} \]