Question

If $n$ is a positive integer, prove that $\sum _{r=1}^n{r}^3{\left(\frac{{}^n{C}_r}{{}^n{C}_{r-1}}\right)}^2=\frac{n{(n+1)}^2(n+2)}{12}$.

Answer 1

${}^n{C}_r=\frac{n!}{(n-r)!.r!}$

$=\frac{(n-(r-1))n!}{(n-(r-1))(n-r)!.(r-1)!.r}$

$=\frac{(n-(r-1))n!}{(n-(r-1)!)(r-1)!.r}$

$=\frac{n-(r-1)}{r}{}^n{C}_{r-1}$

Hence, $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n+1-r}{r}$

Hence, ${\left(\frac{{}^n{C}_r}{{}^n{C}_{r-1}}\right)}^2=\frac{{(n+1-r)}^2}{r^2}$

$r^3{\left(\frac{{}^n{C}_r}{{}^n{C}_{r-1}}\right)}^2=r{(n+1-r)}^2$

$=r[{(n+1)}^2-2r(n+1)+r^2]$

$=r{(n+1)}^2-2r^2(n+1)+r^3$

$\sum r^3{\left(\frac{{}^n{C}_r}{{}^n{C}_{r-1}}\right)}^2=\sum r{(n+1)}^2-2r^2(n+1)+r^3$

$={(n+1)}^2\sum r-2(n+1)\sum r^2+\sum r^3$

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

$=\frac{n{(n+1)}^3}{2}-\frac{2n{(n+1)}^2(2n+1)}{6}+\frac{n^2{(n+1)}^2}{4}$

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

$=\frac{n{(n+1)}^2}{2}\left[\frac{6n+6-8n-4+3n}{6}\right]$

$=\frac{n{(n+1)}^2}{2}\left[\frac{n+2}{6}\right]$

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