If `C_(0),C_(1),C_(2),…,C_(n)` are the binomial coefficients in the expansion of `(1+x)^(n)*n` being even, then `C_(0)+(C_(0)+C_(1))+(C_(0)+C_(1)+C_(2))+….+(C_(0)+C_(1)+C_(2)+…+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 step by step, we need to analyze the expression given and use properties of binomial coefficients.

### Step 1: Understand the expression
We need to find the value of:
\[ C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + C_2 + \ldots + C_{n-1}) \]
This can be rewritten as:
\[ C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + C_2 + \ldots + C_{n-1}) \]

### Step 2: Break down the sums
We can express this sum in terms of binomial coefficients:
- The first term is \(C_0\).
- The second term is \(C_0 + C_1\).
- The third term is \(C_0 + C_1 + C_2\).
- Continuing this way, the last term is \(C_0 + C_1 + C_2 + \ldots + C_{n-1}\).

### Step 3: Count the occurrences of each coefficient
Each coefficient \(C_k\) appears in multiple sums:
- \(C_0\) appears \(n\) times.
- \(C_1\) appears \(n-1\) times.
- \(C_2\) appears \(n-2\) times.
- Continuing this way, \(C_{n-1}\) appears once.

### Step 4: Write the total sum
We can express the total sum as:
\[ nC_0 + (n-1)C_1 + (n-2)C_2 + \ldots + 1C_{n-1} \]

### Step 5: Use the binomial theorem
From the binomial theorem, we know:
\[ (1+x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n \]
Setting \(x = 1\):
\[ 2^n = C_0 + C_1 + C_2 + \ldots + C_n \]
Setting \(x = -1\):
\[ 0 = C_0 - C_1 + C_2 - C_3 + \ldots + (-1)^n C_n \]

### Step 6: Calculate the sums
The sum \(C_0 + C_1 + C_2 + \ldots + C_{n-1}\) can be calculated as:
\[ C_0 + C_1 + C_2 + \ldots + C_{n-1} = 2^n - C_n \]
And \(C_n\) can be expressed as \(C_n = \binom{n}{n} = 1\) when \(n\) is even.

### Step 7: Final expression
Thus, the total sum becomes:
\[ n \cdot 2^n - (n-1) \cdot 2^{n-1} \]
This simplifies to:
\[ n \cdot 2^{n-1} \]

### Conclusion
The final answer is:
\[ n \cdot 2^{n-1} \]