Question

If ${(1+x)}^n=\sum _{r=0}^n{C}_r{x}^r\forall nϵN$ then $\sum _{k=1}^n{k}^3{\left(\frac{C_k}{C_{k-1}}\right)}^2$ equals

• A. $\frac{n(n+1)(n+2)}{12}$
• B. $\frac{n{(n+1)}^2(n+2)}{12}$
• C. $\frac{{(n+1)}^2(n+2)}{12}$
• D. None of these

Answer 1

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

$\frac{C_k}{C_{k-1}}=\frac{n-k+1}{k}$
The required answer=${\sum }_{k=1}^{n}k\times {(n-k+1)}^2$ =$\left({\sum }_{r=1}^n({(n+1)}^2\times k)\right)+({\sum }_{r=1}^n{k}^3)-{\sum }_{r=1}^n\left(\right(2n+2)\times k)$
$=\frac{n\times {(n+1)}^2\times (n+2)}{12}$