Single Choice

The value of $$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$$ is equal to:

A$$680$$
Correct Answer
B$$1085$$
C$$560$$
D$$1240$$

Solution

$$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$$

$$\Rightarrow \left (\dfrac {^{n}C_{r}}{^{n}C_{r - 1}}\right ) = \dfrac{n-r+1}{r}$$

$$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right ) = \displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {n-r+1}{r}\right )\\$$

$$\sum_{r = 1}^{15} (nr-r^2+r) = \sum_{r = 1}^{15} ((n+1)r-r^2)$$

$$ \sum_{r = 1}^{15} (nr-r^2+r) = \dfrac{n(n+1)^2}{2}- \dfrac{n(n+1)(2n+1)}{6}$$

Put $$n=15,$$ we get

$$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right ) = 680$$

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