If n is a positive integer and `C_(k)=""^(n)C_(k)`, then the value of `sum_(k=1)^(n)k^(3)((C_(k))/(C_(k-1)))^(2)` is :

Options

A. `(n+(n+2)(n+1)^(2))/(12)`
B. `(n+(n+1)(n+2)^(2))/(12)`
C. `(n+(n+2)(n+1))/(12)`
D. None of these

Answer

To solve the problem, we need to evaluate the expression:

\[ \sum_{k=1}^{n} k^3 \left( \frac{C_k}{C_{k-1}} \right)^2 \]

where \( C_k = \binom{n}{k} \).

### Step 1: Simplify the Binomial Coefficient Ratio
First, we know that:

\[ \frac{C_k}{C_{k-1}} = \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k} \]

Thus, we can write:

\[ \left( \frac{C_k}{C_{k-1}} \right)^2 = \left( \frac{n-k+1}{k} \right)^2 \]

### Step 2: Substitute into the Summation
Substituting this into our original summation gives:

\[ \sum_{k=1}^{n} k^3 \left( \frac{n-k+1}{k} \right)^2 = \sum_{k=1}^{n} k^3 \cdot \frac{(n-k+1)^2}{k^2} \]

This simplifies to:

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

### Step 3: Expand the Expression
Now we expand \( (n-k+1)^2 \):

\[ (n-k+1)^2 = n^2 - 2nk + k^2 + 2n - 2k + 1 \]

So, we have:

\[ \sum_{k=1}^{n} k(n^2 - 2nk + k^2 + 2n - 2k + 1) \]

### Step 4: Distribute the Summation
Distributing the summation gives:

\[ n^2 \sum_{k=1}^{n} k - 2n \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k^4 + 2n \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

### Step 5: Use Summation Formulas
We use the following formulas for summations:

1. \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
2. \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
3. \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \)

### Step 6: Substitute the Summation Values
Substituting these values into the expression, we get:

\[ n^2 \cdot \frac{n(n+1)}{2} - 2n \cdot \frac{n(n+1)(2n+1)}{6} + \left( \frac{n(n+1)}{2} \right)^2 + 2n \cdot \frac{n(n+1)}{2} - 2 \cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \]

### Step 7: Combine and Simplify
After substituting and simplifying, we can combine like terms. The final expression will yield:

\[ \frac{n(n+1)^2(n+2)}{12} \]

### Final Answer
Thus, the value of the summation is:

\[ \frac{n(n+1)^2(n+2)}{12} \]