Question

If $n$ is positive integer and $k$ is a positive integer not exceeding $n$, then show $\sum _{k=1}^n{k}^3{\left(\frac{{}^n{C}_k}{{}^n{C}_{k-1}}\right)}^2=\frac{n{(n+1)}^2(n+2)}{12}$.

Answer 1

We know that $\frac{C_k}{C_{k-1}}=\frac{{}^n{C}_k}{{}^n{C}_{k-1}}=\frac{n-k+1}{k}$

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

Put $n-k+1=p\Rightarrow k=n-p+1$

when $k=1,p=n$ and when $k=n,p=1$

$\therefore$ Series $\sum _{p=1}^n{p}^2(n-p+1)=\sum _{p=1}^n(n{p}^2-p^3+p^2)=\sum _{p=1}^n(n+1)p^2-\sum _{p=1}^n{p}^3$

$=(n+1)[1^2+2^2+3^2+..+n^2]-[1^3+2^3+3^3+..+n^3]$

$=\frac{(n+1)n(n+1)(2n+1)}{6}-\frac{n^2{(n+1)}^2}{4}$

$=\frac{n{(n+1)}^2}{2}[\frac{2n+1}{3}-\frac{n}{2}]=\frac{n{(n+1)}^2(n+2)}{12}$