0
$\begingroup$

I'm trying to find the value of: $$\sum_{k=0}^{\left \lfloor \frac{p}{2} \right \rfloor} \binom{p}{k}$$ For even and odd $p$, the indication I was given suggests writing it as $$\frac{1}{2}\left (\sum_{k=0}^{\left \lfloor \frac{p}{2} \right \rfloor} \binom{p}{k} + \sum_{k=0}^{\left \lfloor \frac{p}{2} \right \rfloor} \binom{p}{p-k} \right )$$ But I nothing I did got me anywhere.

$\endgroup$
  • 1
    $\begingroup$ Recall that $\sum_{k=0}^p\binom{p}{k}=(1+1)^k$. $\endgroup$ – Robert Z Nov 17 '18 at 15:57
  • $\begingroup$ Look at Pascal's triangle. The p-th row consists of the values $p \choose k$. Maybe you can notice a pattern when you sum 1/2 the values in a row. $\endgroup$ – Joel Pereira Nov 17 '18 at 16:02
0
$\begingroup$

You can write $$ \eqalign{ & 2^{\,p} = \sum\limits_{0\, \le \,k\, \le \,p} {\left( \matrix{ p \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} + \sum\limits_{\left\lfloor {{p \over 2}} \right\rfloor + 1\, \le \,k\, \le \,p} {\left( \matrix{ p \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} + \sum\limits_{\left\lfloor {{p \over 2}} \right\rfloor + 1\, \le \,k\, \le \,p} {\left( \matrix{ p \cr p - k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} + \sum\limits_{0\, \le \,k\, \le \,p - \left\lfloor {{p \over 2}} \right\rfloor - 1} {\left( \matrix{ p \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} + \sum\limits_{0\, \le \,k\, \le \,\left\lceil {{p \over 2}} \right\rceil - 1} {\left( \matrix{ p \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,\left\lceil {{p \over 2}} \right\rceil - 1} {\left( \matrix{ p \cr k \cr} \right)} + \sum\limits_{\left\lceil {{p \over 2}} \right\rceil \, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} + \sum\limits_{0\, \le \,k\, \le \,\left\lceil {{p \over 2}} \right\rceil - 1} {\left( \matrix{ p \cr k \cr} \right)} = \cr & = 2\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} + \left( {\left\lceil {{p \over 2}} \right\rceil - \left\lfloor {{p \over 2}} \right\rfloor - 1} \right)\left( \matrix{ p \cr \left\lfloor {{p \over 2}} \right\rfloor \cr} \right) = \cr & = 2\sum\limits_{0\, \le \,k\, \le \,\left\lfloor {{p \over 2}} \right\rfloor } {\left( \matrix{ p \cr k \cr} \right)} - \left( {1 - p\bmod 2} \right)\left( \matrix{ p \cr \left\lfloor {{p \over 2}} \right\rfloor \cr} \right) \cr} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.