If `(1+x)^(n) =C_(0) +C_(1)x +C_(2)x^(2) +…+C_(n)x^(n)`, then `C_(0) +3C_(1) +5C_(2)+ ….+ (2n+1)C_(n)=`

Options

A. `n.2^(n-1)`
B. `n.2^(n)`
C. `(n+1) 2^(n)`
D. none

Answer

To solve the problem \( C_0 + 3C_1 + 5C_2 + \ldots + (2n+1)C_n \) where \( (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \), we will follow these steps:

### Step 1: Substitute \( x \) with \( y^2 \) and multiply by \( y \)

We start by substituting \( x \) with \( y^2 \) in the binomial expansion:

\[ (1 + y^2)^n = C_0 + C_1 y^2 + C_2 y^4 + \ldots + C_n y^{2n} \]

Now, multiply the entire equation by \( y \):

\[ y(1 + y^2)^n = C_0 y + C_1 y^3 + C_2 y^5 + \ldots + C_n y^{2n + 1} \]

### Step 2: Differentiate both sides with respect to \( y \)

Next, we differentiate both sides with respect to \( y \):

\[ \frac{d}{dy} \left[ y(1 + y^2)^n \right] = \frac{d}{dy} \left[ C_0 y + C_1 y^3 + C_2 y^5 + \ldots + C_n y^{2n + 1} \right] \]

Using the product rule on the left side, we get:

\[ (1 + y^2)^n + y \cdot n(1 + y^2)^{n-1} \cdot 2y = C_0 + 3C_1 y^2 + 5C_2 y^4 + \ldots + (2n + 1)C_n y^{2n} \]

This simplifies to:

\[ (1 + y^2)^n + 2ny^2(1 + y^2)^{n-1} = C_0 + 3C_1 y^2 + 5C_2 y^4 + \ldots + (2n + 1)C_n y^{2n} \]

### Step 3: Set \( y = 1 \)

Now, we substitute \( y = 1 \):

\[ (1 + 1^2)^n + 2n(1^2)(1 + 1^2)^{n-1} = C_0 + 3C_1 + 5C_2 + \ldots + (2n + 1)C_n \]

This gives us:

\[ 2^n + 2n \cdot 1 \cdot 2^{n-1} = C_0 + 3C_1 + 5C_2 + \ldots + (2n + 1)C_n \]

### Step 4: Simplify the expression

Simplifying the left side:

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

### Final Result

Thus, we find that:

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

### Summary

The final result is:

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