If `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+…+C_(n)x^(n)`, then `C_(0)+5C_(1)+9C_(2)+ …+(4n+1)C_(n)=`

Options

A. `n.2^(n)`
B. `(n+1)2^(n)`
C. `(2n+1)2^(n)`
D. `(4n+1)2^(n)`

Answer

To solve the problem, we need to evaluate the expression \( C_0 + 5C_1 + 9C_2 + \ldots + (4n + 1)C_n \) given that \( (1+x)^n = C_0 + C_1x + C_2x^2 + \ldots + C_nx^n \).

### Step-by-Step Solution:

1. **Substitution**:
We start by substituting \( x = y^2 \) in the binomial expansion:
\[ (1 + y^2)^n = C_0 + C_1y^2 + C_2y^4 + \ldots + C_ny^{2n} \]
This gives us:
\[ y(1 + y^4)^n = C_0y + C_1y^3 + C_2y^5 + \ldots + C_ny^{2n + 1} \]

2. **Differentiation**:
Next, we differentiate both sides with respect to \( y \):
\[ \frac{d}{dy}[y(1 + y^4)^n] = (1 + y^4)^n + y \cdot n(1 + y^4)^{n-1} \cdot 4y^3 \]
The left-hand side becomes:
\[ (1 + y^4)^n + 4ny^4(1 + y^4)^{n-1} \]

3. **Evaluating at \( y = 1 \)**:
Now we substitute \( y = 1 \):
\[ (1 + 1^4)^n + 4n \cdot 1^4(1 + 1^4)^{n-1} = 2^n + 4n \cdot 2^{n-1} \]
This simplifies to:
\[ 2^n + 4n \cdot 2^{n-1} = 2^n + 2n \cdot 2^n = (1 + 2n)2^n \]

4. **Final Result**:
Thus, we find that:
\[ C_0 + 5C_1 + 9C_2 + \ldots + (4n + 1)C_n = (1 + 2n)2^n \]

### Conclusion:
The value of \( C_0 + 5C_1 + 9C_2 + \ldots + (4n + 1)C_n \) is \( (1 + 2n)2^n \).