Evaluating a Series of Binomial Co-efficients

**Stephen Shaw** · February 19th 2009, 03:39 AM

Hey guys,

I've been scratching my head over this one:

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

That is to say,

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

My first thought was to say

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

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

**Plato** · February 19th 2009, 04:46 AM

Originally Posted by Stephen Shaw

Evaluate:
$\binom{40}{0}+\binom{40}{1}+\binom{40}{2}+\dots+\b inom{40}{40}$
That is to say,
$\sum_{n=0}^{40}\binom{40}{n}$

Think about it for a moment.
That is simply the number of subsets of a set of 40 elements: $2^{40}$ .
Another way. $\left( {x + y} \right)^{40} = \sum\limits_{k = 0}^{40} {{40 \choose k}x^{40 - k} y^k }$
Let $x=1\;\&\;y=1$ we have $\left( {2} \right)^{40} = \sum\limits_{k = 0}^{40} {{40 \choose k} }$

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