Question

The value of ${\sum }_{r=1}^{15}{r}^2\left(\frac{{}^{15}Cr}{{}^{15}Cr-1}\right)$ is equal to:

Answer 1

$\sum _{r=1}^{15}{r}^2\left(\frac{{}^{15}{C}_r}{{}^{15}{C}_{r-1}}\right)$

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

$\frac{{}^{15}{C}_r}{{}^{15}{C}_{r-1}}=\frac{15-r+1}{r}=\frac{16-r}{r}$

$\sum _{r=1}^{15}{r}^2\left(\frac{{}^{15}{C}_r}{{}^{15}{C}_{r-1}}\right)=\sum _{r=1}^{15}{r}^2\left(\frac{16-r}{r}\right)$

$=\sum _{r=1}^{15}16r-r^2$

$=16\sum _{r=1}^{15}r-\sum _{r=1}^{15}{r}^2$

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

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

$=16\left(\frac{15(15+1)}{2}\right)-\frac{15(15+1)\left(2\right(15)+1)}{6}$

$=15\left(16\right)[8-\frac{31}{6}]=15\left(16\right)\frac{17}{6}$

$=5\times 8\times 17=40\times 17$

$=680$.