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, we need to evaluate the expression:

\[ S = C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + C_2 + \ldots + C_{n-1}) \]

where \( C_r \) are the binomial coefficients from the expansion of \( (1+x)^n \).

### Step 1: Rewrite the Expression
The expression can be rewritten as:

\[ S = nC_0 + (n-1)C_1 + (n-2)C_2 + \ldots + 1C_{n-1} \]

This is because \( C_0 \) appears \( n \) times, \( C_1 \) appears \( n-1 \) times, and so on, until \( C_{n-1} \) which appears once.

### Step 2: Generalize the Expression
We can express \( S \) in summation notation:

\[ S = \sum_{r=0}^{n-1} (n-r) C_r \]

### Step 3: Split the Summation
We can split this summation into two parts:

\[ S = n \sum_{r=0}^{n-1} C_r - \sum_{r=0}^{n-1} r C_r \]

### Step 4: Evaluate the First Summation
The first summation \( \sum_{r=0}^{n-1} C_r \) is the sum of the binomial coefficients from \( C_0 \) to \( C_{n-1} \). From the binomial theorem, we know:

\[ \sum_{r=0}^{n} C_r = 2^n \]

Thus,

\[ \sum_{r=0}^{n-1} C_r = 2^n - C_n = 2^n - 1 \]

### Step 5: Evaluate the Second Summation
The second summation \( \sum_{r=0}^{n-1} r C_r \) can be evaluated by differentiating the binomial expansion:

\[ \sum_{r=0}^{n} C_r x^r = (1+x)^n \]

Differentiating both sides with respect to \( x \):

\[ \sum_{r=1}^{n} r C_r x^{r-1} = n(1+x)^{n-1} \]

Setting \( x = 1 \):

\[ \sum_{r=1}^{n} r C_r = n \cdot 2^{n-1} \]

Thus,

\[ \sum_{r=0}^{n-1} r C_r = n \cdot 2^{n-1} - C_n = n \cdot 2^{n-1} - 1 \]

### Step 6: Substitute Back into the Expression for S
Now substituting back into our expression for \( S \):

\[ S = n(2^n - 1) - (n \cdot 2^{n-1} - 1) \]

### Step 7: Simplify the Expression
Simplifying \( S \):

\[ S = n \cdot 2^n - n - n \cdot 2^{n-1} + 1 \]

Factoring out \( n \):

\[ S = n \cdot 2^{n-1} (2 - 1) + 1 - n = n \cdot 2^{n-1} + 1 - n \]

### Final Result
Thus, the final result is:

\[ S = n \cdot 2^{n-1} + 1 - n \]