Question

If $n$ is a positive integer and $C_k={{}^{n}C}_k$, find the value of ${\sum }_{k=1}^n{k}^3{\left(\frac{C_k}{C_{k-1}}\right)}^2$

• A. $\frac{n{(n-1)}^2(n+2)}{12}$
• B. $\frac{n{(n+1)}^2(n-2)}{12}$
• C. $\frac{n{(n+1)}^2(n+2)}{12}$
• D. $\frac{n{(n-1)}^2(n-2)}{12}$

Answer 1

The correct option is C $\frac{n{(n+1)}^2(n+2)}{12}$

$\sum _{k=1}^n{k}^3{\left(\frac{C_k}{C_{k-1}}\right)}^2=\sum _{k=1}^n{k}^3{\left(\frac{n-k+1}{k}\right)}^2$

$={\sum }_{k=1}^{n}k{(n-k+1)}^2$

$={(n+1)}^2{\sum }_{k=1}^{n}k+{\sum }_{k=1}^n{k}^3-2(n+1){\sum }_{k=1}^n{k}^2$
$=\frac{n{(n+1)}^3}{2}+\frac{n^2{(n+1)}^2}{4}-\frac{n{(n+1)}^2(2n+1)}{3}$

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