Question

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

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

Answer 1

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

$\sum _{k=1}^n{k}^3{\left(\frac{C_k}{C_{k-1}}\right)}^2=\sum _{k=1}^n{k}^3{\left(\frac{n-k+1}{k}\right)}^2$
$$[\because \frac{{}^n{C}_k}{{}^n{C}_{k-1}}=\frac{n-k+1}{k}]$$
$=\sum _{k=1}^{n}k(n-k+1{)}^2=\sum _{k=1}^{n}k\left[\right(n+1{)}^2-2k(n+1)+k^2]=(n+1{)}^2\sum _{k=1}^{n}k-2(n+1)\sum _{k=1}^n{k}^2+\sum _{k=1}^n{k}^3=(n+1{)}^2\cdot \frac{n(n+1)}{2}-2(n+1)\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n^2(n+1{)}^2}{4}=\frac{n(n+1{)}^2}{12}\left[6\right(n+1)-4(2n+1)+3n]=\frac{n(n+1{)}^2}{12}\cdot (n+2)=\frac{n(n+2)(n+1{)}^2}{12}$