If n is a positive integer and Ck=(nk), then C_k = {n \choose k}, $$\text{ then}$$ \\[6pt]Ck=(kn), then ∑k=1nk3(CkCk−1)2 equals:\sum_{k=1}^{n} k^3 \left( $$\frac{C_k}$${C_{k-1}} \right)^2 $$\text{ equals:}$$ \\[10pt]k=1∑nk3(Ck−1Ck)2 equals:

112 n(n+1)(n+2)$$\frac{1}{12}$$ \, n(n+1)(n+2) \\[6pt]121n(n+1)(n+2)

112 n(n+1)2(n+2)$$\frac{1}{12}$$ \, n(n+1)^2(n+2) \\[6pt]121n(n+1)2(n+2)

112 n(n+1)(n+2)$$2\frac{1}{12}$$ \, n(n+1)(n+2)^2 \\[6pt]121n(n+1)(n+2)2

112 n2(n+1)(n+2)$$\frac{1}{12}$$ \, n^2(n+1)(n+2) \\[12pt]121n2(n+1)(n+2)

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If  n is a positive integer and Ck=(nk), then C_k = {n \choose k}, \text{ then} \\[6pt]Ck=(kn), then  ∑k=1nk3(CkCk−1)2

Question

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

$\frac{1}{12} \, n(n+1)(n+2) \\[6pt]$

$\frac{1}{12} \, n(n+1)^2(n+2) \\[6pt]$

$\frac{1}{12} \, n(n+1)(n+2)^2 \\[6pt]$

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

Solution

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)