If `C_(0),C_(1), C_(2),...,C_(n)` denote the cefficients in
the expansion of `(1 + x)^(n)`, then
`C_(0) + 3 .C_(1) + 5 . C_(2)+ ...+ (2n + 1) C_(n) = ` .

Options

A. `2^(n)`
B. `2^(n)+n*2^(n-1)`
C. `2^(n)*(n+1)`
D. None of these

Answer

To solve the problem, we need to find the value of the sum:

\[ S_n = C_0 + 3C_1 + 5C_2 + \ldots + (2n + 1)C_n \]

where \( C_r = \binom{n}{r} \) are the coefficients in the expansion of \( (1 + x)^n \).

### Step-by-step Solution:

1. **Express the sum in terms of binomial coefficients:**
\[ S_n = \sum_{r=0}^{n} (2r + 1) C_r = \sum_{r=0}^{n} (2r + 1) \binom{n}{r} \]

2. **Separate the sum into two parts:**
\[ S_n = \sum_{r=0}^{n} 2r \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r} \]

3. **Evaluate the second sum:**
The sum of the binomial coefficients is:
\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

4. **Evaluate the first sum using the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \):**
\[ \sum_{r=0}^{n} r \binom{n}{r} = n \sum_{r=1}^{n} \binom{n-1}{r-1} = n \cdot 2^{n-1} \]

5. **Substituting back into the equation for \( S_n \):**
\[ S_n = 2 \cdot n \cdot 2^{n-1} + 2^n \]

6. **Simplifying the expression:**
\[ S_n = 2n \cdot 2^{n-1} + 2^n = 2^{n} (n + 1) \]

Thus, the final result is:
\[ S_n = 2^n (n + 1) \]

### Final Answer:
\[ S_n = 2^n (n + 1) \]