Hey guys,

I've been scratching my head over this one:

Evaluate:

$\displaystyle \binom{40}{0}+\binom{40}{1}+\binom{40}{2}+\dots+\b inom{40}{40}$

That is to say,

$\displaystyle \sum_{n=0}^{40}\binom{40}{n}$

My first thought was to say

$\displaystyle \sum_{n=0}^{40}\binom{40}{n} = \sum_{n=0}^{40}\frac{40!}{40(40-n)!} = \sum_{n=0}^{40}\left(\frac{A}{40}+\frac{B}{(40-n)!}\right)$

Then to find $\displaystyle A$ and $\displaystyle B$ by partial fractions and do a summation table, but couldn't get it out. Any suggestions from anyone?