If n is a positive integer and C_k = {^nC_k}, | vedclass

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(D) We know that $\frac{C_k}{C_{k - 1}} = \frac{n - k + 1}{k}$.Substituting this into the expression,we get:$\sum\limits_{k = 1}^n {k^3\left( {\frac{n

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

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(D) We know that $\frac{C_k}{C_{k - 1}} = \frac{n - k + 1}{k}$.Substituting this into the expression,we get:$\sum\limits_{k = 1}^n {k^3\left( {\frac{n - k + 1}{k}} \right)^2} = \sum\limits_{k = 1}^n {k(n - k + 1)^2}$$= \sum\limits_{k = 1}^n {k((n + 1) - k)^2} = \sum\limits_{k = 1}^n {k((n + 1)^2 - 2k(n + 1) + k^2)}$$= (n + 1)^2 \sum\limits_{k = 1}^n k - 2(n + 1) \sum\limits_{k = 1}^n k^2 + \sum\limits_{k = 1}^n k^3$$= (n + 1)^2 \frac{n(n + 1)}{2} - 2(n + 1) \frac{n(n + 1)(2n + 1)}{6} + \frac{n^2(n + 1)^2}{4}$$= \frac{n(n + 1)^2}{2} \left[ (n + 1) - \frac{2(2n + 1)}{3} + \frac{n}{2} \right]$$= \frac{n(n + 1)^2}{2} \left[ \frac{6n + 6 - 8n - 4 + 3n}{6} \right]$$= \frac{n(n + 1)^2}{12} (n + 2)$.

If $n$ is a positive integer and $C_k = {^nC_k}$,then the value of $\sum\limits_{k = 1}^n {k^3\left( {\frac{C_k}{C_{k - 1}}} \right)^2}$ is:

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