XAT 2014 — QA & DI Question 22

$\sum _{i=2}^{100}\frac{1}{{\log }_{i}100!}=\sum _{i=2}^{100}{\log }_{100!}2+{\log }_{100!}3+{\log }_{100!}4+...+{\log }_{100!}100$

$={\log }_{100!}100!=1$

Hence, option (c).