If `(1+x)^n = C_0 + C_1 x + C_2x^2 + …………+C_n x^n` , find the value of `C_0 - 2C_1 + 3C_2 - ……….+ (-1)^n (n+1) C_n`

Answer

To find the value of \( C_0 - 2C_1 + 3C_2 - \ldots + (-1)^n (n+1) C_n \), we start from the binomial expansion of \( (1+x)^n \):

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

### Step 1: Multiply the entire equation by \( x \)

We multiply both sides of the equation by \( x \):

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

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

Now we differentiate both sides with respect to \( x \):

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

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

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

### Step 3: Substitute \( x = -1 \)

Next, we substitute \( x = -1 \) into the differentiated equation:

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

This simplifies to:

\[ 0 + 0 = C_0 - 2C_1 + 3C_2 - \ldots + (-1)^n (n+1) C_n \]

### Step 4: Evaluate the left side

The left side equals \( 0 \), so we have:

\[ C_0 - 2C_1 + 3C_2 - \ldots + (-1)^n (n+1) C_n = 0 \]

### Final Result

Thus, the value of \( C_0 - 2C_1 + 3C_2 - \ldots + (-1)^n (n+1) C_n \) is:

\[ \boxed{0} \]