New user? Sign up

Existing user? Log in

$\displaystyle \sum_{n=3}^{\infty} \dfrac{1}{\binom{n}{3}} = \dfrac{1}{\binom{3}{3}} + \dfrac{1}{\binom{4}{3}} + \dfrac{1}{\binom{5}{3}} + \dfrac{1}{\binom{6}{3}} + \cdots = \, ?$

Problem Loading...

Note Loading...

Set Loading...