Question

The value of ${\sum }_{r=1}^n{r}^2$$\frac{{}^n{C}_r}{{}^n{C}_{r-1}}$is

• A. $\frac{n(n+1)(n+2)}{12}$
• B. $\frac{n^2(n+2)}{6}$
• C. $\frac{n(n+1)}{2}$
• D. $\frac{n(n+1)(n+2)}{6}$

Answer 1

The correct option is D $\frac{n(n+1)(n+2)}{6}$

${\sum }_{r=1}^n{r}^2\frac{\stackrel{n}{\underset{r}{C}}}{\stackrel{n}{\underset{r-1}{C}}}$

we know $\stackrel{n}{\underset{r}{C}}=\frac{n!}{(n-r)!r!}$ and

$\stackrel{n}{\underset{r-1}{C}}=\frac{n!}{(n-r+1)!(r-1)!}$

So we divide both we get

$\frac{\stackrel{n}{\underset{r}{C}}}{\stackrel{n}{\underset{r-1}{C}}}=\frac{(n-r+1)}{r}{\sum }_{r=1}^n{r}^2\frac{\stackrel{n}{\underset{r}{C}}}{\stackrel{n}{\underset{r-1}{C}}}={\sum }_{r=1}^n{r}^2\frac{(n-r+1)}{r}={\sum }_{r=1}^{n}r(n+1)-r^2$

We know

$\sum r=\frac{n(n+1)}{2}\sum r^2=\frac{n(n+1)(2n+1)}{6}$

Using this we have

${\sum }_{r=1}^n{r}^2\frac{\stackrel{n}{\underset{r}{C}}}{\stackrel{n}{\underset{r-1}{C}}}=\frac{n{(n+1)}^2}{2}-\frac{n(n+1)(2n+1)}{6}{\sum }_{r=1}^n{r}^2\frac{\stackrel{n}{\underset{r}{C}}}{\stackrel{n}{\underset{r-1}{C}}}=\frac{n(n+1)(n+2)}{6}$