`C_1 + 2. C_2 + 3. C_3 + …... + n. C_n=`

Options

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

Answer

To solve the problem \( C_1 + 2C_2 + 3C_3 + \ldots + nC_n \), we will utilize the Binomial Theorem and differentiation. Here’s a step-by-step solution:

### Step 1: Recall the Binomial Expansion
The Binomial Theorem states that:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + \ldots + C_n x^n \]
where \( C_k \) represents the binomial coefficients.

**Hint:** Remember that \( C_k = \binom{n}{k} \).

### Step 2: Differentiate the Binomial Expansion
To incorporate the coefficients \( 1, 2, 3, \ldots, n \) into our series, we differentiate both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}[(1 + x)^n] = n(1 + x)^{n-1} \]
On the right side, differentiating gives:
\[ C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1} \]

**Hint:** Differentiation will bring down the exponent as a coefficient.

### Step 3: Set \( x = 1 \)
Now, we substitute \( x = 1 \) into the differentiated equation:
\[ n(1 + 1)^{n-1} = C_1 + 2C_2(1) + 3C_3(1^2) + \ldots + nC_n(1^{n-1}) \]
This simplifies to:
\[ n \cdot 2^{n-1} = C_1 + 2C_2 + 3C_3 + \ldots + nC_n \]

**Hint:** Substituting \( x = 1 \) simplifies the equation significantly.

### Step 4: Conclusion
Thus, we find that:
\[ C_1 + 2C_2 + 3C_3 + \ldots + nC_n = n \cdot 2^{n-1} \]

**Final Answer:**
\[ C_1 + 2C_2 + 3C_3 + \ldots + nC_n = n \cdot 2^{n-1} \]