If `C_o, C_1, C_2 ....,C_n`C denote the binomial coefficients in the expansion of `(1 + x)^n,` then `1^3. C_1 + 2^3 . C_2 + 3^3 .C_3 + ... + n^3 .C_n, =`

Options

A. `(n+2)(n+3)3^(n-3)`
B. `n^(2) (n+3) 2^(n-3)`
C. `n^(2) (n+3) 2^(n)`
D. `n(n+ 1)(n+2)2^(n-3)`

Explanation

To solve the problem, we need to find the value of the expression \(1^3 \cdot C_1 + 2^3 \cdot C_2 + 3^3 \cdot C_3 + \ldots + n^3 \cdot C_n\), where \(C_k\) are the binomial coefficients in the expansion of \((1 + x)^n\).

### Step-by-step Solution:

1. **Understand the Binomial Expansion**:
The binomial expansion of \((1 + x)^n\) is given by:
\[ (1 + x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
where \(C_k = \binom{n}{k}\).

2. **Differentiate the Expansion**:
Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}[(1 + x)^n] = n(1 + x)^{n-1} \]
This gives:
\[ C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1} = n(1 + x)^{n-1} \]

3. **Multiply by \(x\)**:
Multiply the entire equation by \(x\):
\[ C_1 x + 2C_2 x^2 + 3C_3 x^3 + \ldots + nC_n x^n = n x(1 + x)^{n-1} \]

4. **Differentiate Again**:
Differentiate this equation again with respect to \(x\):
\[ C_1 + 2 \cdot 2C_2 x + 3 \cdot 3C_3 x^2 + \ldots + n \cdot nC_n x^{n-1} = n \left[(1 + x)^{n-1} + x(n-1)(1 + x)^{n-2}\right] \]

5. **Multiply by \(x\) Again**:
Multiply the equation by \(x\) again:
\[ C_1 x + 2 \cdot 2C_2 x^2 + 3 \cdot 3C_3 x^3 + \ldots + n \cdot nC_n x^n = n x(1 + x)^{n-1} + n(n-1)x^2(1 + x)^{n-2} \]

6. **Differentiate Once More**:
Differentiate this equation once more:
\[ C_1 + 2 \cdot 2C_2 x + 3 \cdot 3C_3 x^2 + \ldots + n \cdot nC_n x^{n-1} = n \left[(1 + x)^{n-1} + x(n-1)(1 + x)^{n-2}\right] \]

7. **Substitute \(x = 1\)**:
Set \(x = 1\) in the final differentiated equation:
\[ C_1 + 2^3 C_2 + 3^3 C_3 + \ldots + n^3 C_n = n \left[(1 + 1)^{n-1} + 1(n-1)(1 + 1)^{n-2}\right] \]
This simplifies to:
\[ C_1 + 2^3 C_2 + 3^3 C_3 + \ldots + n^3 C_n = n \left[2^{n-1} + (n-1)2^{n-2}\right] \]

8. **Final Simplification**:
Combine the terms:
\[ = n \left[2^{n-2}(2 + (n-1))\right] = n \cdot n \cdot 2^{n-2} \]

Thus, the final result is:
\[ 1^3 \cdot C_1 + 2^3 \cdot C_2 + 3^3 \cdot C_3 + \ldots + n^3 \cdot C_n = n^2 \cdot 2^{n-2} \]