The value of r = 1 15 r 2 ( C r 15 C r - 1 15 ) is equal to

Question

The value of $\sum_{r = 1}^{15} r^{2} (\frac{{}^{15}{C_{r}}}{{}^{15}{C_{r - 1}}})$ is equal to

Smart Achievers · Accepted Answer

(4)

We have,

$\frac{{}^{15}C_{r}}{{}^{15}C_{r - 1}} = \frac{15!}{r! (15 - r)!} \times \frac{(r - 1)! (15 - r + 1)!}{15!} = \frac{16 - r}{r}$

$∴$ $\sum_{r = 1}^{15} r^{2} (\frac{{}^{15}C_{r}}{{}^{15}C_{r - 1}}) = \sum_{r = 1}^{15} r^{2} (\frac{16 - r}{r}) = \sum_{r = 1}^{15} (16 r - r^{2})$

$= 16 \times \frac{15 \times 16}{2} - \frac{15 \times 16 \times 31}{6} = 680$

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