JEE MATHEMATICS MCQ: COMBINATIONS

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{\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = \frac{k}{{21}},

then

k

equals

View answer and explanation

Correct option: A

100

Option A (100) is correct because after simplifying the summation using combinatorial identities and the formula for the sum of cubes, we find that k = 100.

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{\;^{n + 2}}{C_8}:{\;^{n - 2}}{P_4} = \frac{{57}}{{16}}

, then

n

is equal to