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Q.
If $\sum\limits^{20}_{i=1} \left( \frac{^{20}C_{i-1}}{^{20}C_{i} + ^{20 }C_{i-1}}\right)^3 = \frac{k }{21} $, then $k$ equals :
JEE MainJEE Main 2019Permutations and Combinations
Solution:
$\sum_{i=1}^{20}\left(\frac{{ }^{20} C _{ i -1}}{{ }^{20} C _{ i }+{ }^{20} C _{ i -1}}\right)^{3}=\frac{ k }{21}$
$\Rightarrow \sum_{i=1}^{20}\left(\frac{{ }^{20} C_{i-1}}{{ }^{21} C_{i}}\right)^{3}=\frac{k}{21}$
$\Rightarrow \sum_{ i =1}^{20}\left(\frac{ i }{21}\right)^{3}=\frac{ k }{21}$
$\Rightarrow \frac{1}{(21)^{3}}\left[\frac{20(21)}{2}\right]^{2}=\frac{ k }{21}$
$\Rightarrow 100= k$