JEE Mains · Maths · STD 11 - 6. permutation and combination
The value of \(\sum\limits_{r = 1}^{15} {{r^2}\,\left( {\frac{{^{15}{C_r}}}{{^{15}{C_{r - 1}}}}} \right)} \) is equal to
- A \(1240\)
- B \(560\)
- C \(1085\)
- D \(680\)
Answer & Solution
Correct Answer
(D) \(680\)
Step-by-step Solution
Detailed explanation
\(\sum\limits_{r = 1}^{15} {{r^2}} \,\left( {\frac{{{{15}_{{C_r}}}}}{{{{15}_{{C_{r - 1}}}}}}} \right)\) \(\frac{{{{15}_{{C_r}}}}}{{{{15}_{C - 1}}}} = \frac{{\frac{{\frac{{15!}}{{15! - rr!}}}}{{15!}}}}{{r - 1!\,16 - r!}}\)…
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