If  n is a positive integer and $$C_k = {n \choose k}, \text{ then} \\[6pt]$$  $$\sum_{k=1}^{n} k^3 \left( \frac{C_k}{C_{k-1}} \right)^2 \text{ equals:} \\[10pt]$$​​

Correct option is C

Given:
$$C_k = {n \choose k}$$​ and we are to evaluate:
​$$\sum_{k=1}^{n} k^3 \left( \frac{{n \choose k}}{{n \choose k-1}} \right)^2 \\$$​​

Concept used:
Use property of binomial coefficients:
​$$\frac{{n \choose k}}{{n \choose k-1}} = \frac{n - k + 1}{k} \\$$​​
Simplify:
​$$\left( \frac{{n \choose k}}{{n \choose k-1}} \right)^2 = \left( \frac{n - k + 1}{k} \right)^2 \\\Rightarrow \sum_{k=1}^{n} k^3 \cdot \left( \frac{n - k + 1}{k} \right)^2 = \sum_{k=1}^{n} (n - k + 1)^2 \cdot k \\$$​​

Let $$r = n - k + 1 \Rightarrow k = n - r + 1 \\$$​​
​$$\Rightarrow \sum_{r=1}^{n} r^2 (n - r + 1)^3 \\$$​​
This is a standard identity result:
​$$\sum_{k=1}^{n} k^3 \left( \frac{{n \choose k}}{{n \choose k-1}} \right)^2 = \frac{1}{12} \, n(n+1)(n+2)^2 \\$$​​
Correct answer is (c)