JEE MATHEMATICS MCQ: COMBINATIONS

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\sum_{i=1}^{20}\left(\frac{{ }^{20} \mathrm{C}_{i-1}}{{ }^{20} \mathrm{C}_{i}+{ }^{20} \mathrm{C}_{i-1}}\right)^{3}=\frac{k}{21}

, then

k

equals:

View answer and explanation

Correct option: D

100

The correct answer is D (100) because after simplifying the expression inside the summation and using the formula for the sum of cubes of first n natural numbers, we get k = 100.

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

, then

n

is equal to