Use the general/expanded binomial theorem i.e. $ \displaystyle{{m \choose n} = \frac {m!}{n!(m - n)!} = \frac{m (m - 1) (m - 2) (m - 3) \cdot \cdot \cdot (m - n + 1)}{n!}}$ Then apply it to $ (a + b)^{\sqrt {2}}$ EDIT: wow, i need to type slower

falconwing64 wrote: $ \displaystyle{{m \choose n} = \frac {m!}{(m - n)!} = m (m - 1) (m - 2) (m - 3) \cdot \cdot \cdot (m - n + 1)}$ $ \displaystyle{{m \choose n} = \frac {m!}{(m - n)!n!}}$

Easier solution: Writed $ \sqrt 2$ in basis $ 2$ and we can can compute every number of form $ 2^{\frac{m}{2^n}}$.