Mathematics - Binomial Theorem Question with Solution | TestHub

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

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

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

$={\displaystyle \sum _{r=1}^{15}r}$$\left(16-r\right)$

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

$=$$\frac{16 \times 15 \times 16}{2}-\frac{15 \times 31 \times 16}{6}$

$=$$8 \times 15 \times 16-5 \times 8 \times 31$

$=$$1920-1240=680$