Sum it

Algebra Level 4

If \[ \large\displaystyle \sum^{99}_{k=0}\dbinom{99}{k}\] is divided by \[ \large\displaystyle\sum^{33}_{k=0}\dbinom{99}{3k}, \]
the resulting number is of the form \(\large{\dfrac{a}{b}} \dot \large {\dfrac{c^d}{(e^f-1)}},\) where \(\large{a, b, c, e}\) are prime numbers. Evaluate \(\large{a+b+c+d+e+f-1}\).

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