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. (a) `(n (n+2)(n+1)^(2))/(12)`
B. (b) `(n(n+1)(n+2)^(2))/(12)`
C. (c) `(n(n+2)(n+1))/(12)`
D. (d) None of these

Answer

To solve the given 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: Understanding the terms \( C_k \) and \( C_{k-1} \)

The binomial coefficient \( C_k \) is defined as:

\[ C_k = \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

Similarly, we can express \( C_{k-1} \):

\[ C_{k-1} = \binom{n}{k-1} = \frac{n!}{(k-1)!(n-k+1)!} \]

### Step 2: Finding the ratio \( \frac{C_k}{C_{k-1}} \)

Now, we compute the ratio:

\[ \frac{C_k}{C_{k-1}} = \frac{\frac{n!}{k!(n-k)!}}{\frac{n!}{(k-1)!(n-k+1)!}} = \frac{(n-k+1)}{k} \]

### Step 3: Squaring the ratio

Next, we square this ratio:

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

### Step 4: Substituting back into the summation

Now we substitute this back into our summation:

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

### Step 5: Expanding the expression

We can expand \( (n-k+1)^2 \):

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

Thus, we have:

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

### Step 6: Distributing the summation

We can distribute the summation:

\[ = n^2 \sum_{k=1}^{n} k - 2n \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k^3 + 2n \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k \]

### Step 7: Using summation formulas

Using the formulas for the 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 8: Putting it all together

Now we substitute these formulas into our expression and simplify. After simplification, we will find:

\[ \sum_{k=1}^{n} k(n-k+1)^2 = \frac{n(n+1)(n+2)(n+1)}{12} \]

### Final Result

Thus, the value of the original summation is:

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

This corresponds to option A:

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