The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to
The value of $\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} $ is equal to
- [JEE MAIN 2016]
- A
$1240$
- B
$560$
- C
$1085$
- D
$680$
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- [IIT 1980]
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- [JEE MAIN 2020]
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