If

Consider the expression, $\frac{{ }^{20} C_{i-1}}{{ }^{20} C_{i}+{ }^{20} C_{i-1}}=\frac{{ }^{20} C_{i-1}}{{ }^{21} C_{1}}$

$=\frac{20 !}{(i-1) !(21-i) !} \times \frac{i !(21-i) !}{21 !}=\frac{i}{21}$

$\therefore \quad \sum_{i=1}^{20}\left(\frac{{ }^{20} C_{i-1}}{{ }^{20} C_{i}+{ }^{20} C_{i-1}}\right)^{3}$

$=\sum_{i=1}^{20}\left(\frac{i}{21}\right)^{3}=\frac{(1)}{(21)^{3}} \sum_{i=1}^{20} i^{3}$

$=\frac{1}{(21)^{3}} \times\left(\frac{20 \times 21}{2}\right)^{2}=\frac{100}{21}$

$\because \sum_{i=1}^{20}\left(\frac{{ }^{20} C_{i-1}}{{ }^{20} C_{i}+{ }^{20} C_{i-1}}\right)^{3}=\frac{k}{21}$

$\therefore \quad k=100$