Sum of a combination












0












$begingroup$


i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations







share|cite|improve this question









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  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14
















0












$begingroup$


i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14














0












0








0





$begingroup$


i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations







share|cite|improve this question









$endgroup$




i cant find an expression for:



$sum_{k=0}^{lfloor
frac{p}{2}rfloor}$ ${p}choose{k}$



The hint given by the exercise is to do it by seperating the cases where p is odd and even.



What I did so far:
When $p$ is even, $lfloor p/2 rfloor = p/2$, so all we need is find an expression
$sum_{k=0}^{ frac{p}{2}}$ ${p}choose{k}$, which I was not able to do.



When $p$ is odd, there is $q$ so that $p=2q+1$ and $p/2= q +1/2$ which means $lfloor p/2 rfloor = q$, and I really dont know what to do with this.



The previous question was to show that functions that are of the form $ax+b$ are both convex and concave.



I thought that I should try to put some values in to look for a pattern and here is what I have :
for $p$ even :



$p=2, sum_{k=0}^{1}$ ${2}choose{k}$$=1+2=3$



$p=4, sum_{k=0}^{2}$ ${4}choose{k}$$=1+6+4=11$



$p=6, sum_{k=0}^{3}$ ${6}choose{k}$$=1+6+15+20=42$



no apparent pattern.



when p is odd,



$p=1, sum_{k=0}^{0}$ ${1}choose{k}$$=1$



$p=3, sum_{k=0}^{1}$ ${3}choose{k}$$=1+3=4$



$p=5, sum_{k=0}^{2}$ ${5}choose{k}$$=1+5+10=16$



$p=7, sum_{k=0}^{4}$ ${7}choose{k}$$=1+7+21+35+35=99$



No apparent pattern here



Thank you for your help






summation combinations




summation combinations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 13 at 20:52









user310148user310148

43




43








  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14














  • 1




    $begingroup$
    Your calculation for $p=7$ is wrong.
    $endgroup$
    – Robert Israel
    Jan 13 at 21:14








1




1




$begingroup$
Your calculation for $p=7$ is wrong.
$endgroup$
– Robert Israel
Jan 13 at 21:14




$begingroup$
Your calculation for $p=7$ is wrong.
$endgroup$
– Robert Israel
Jan 13 at 21:14










1 Answer
1






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oldest

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1












$begingroup$

We have ${p choose k} = {p choose p-k}$, so
$$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






share|cite|improve this answer









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    $begingroup$

    We have ${p choose k} = {p choose p-k}$, so
    $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
    Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
    If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We have ${p choose k} = {p choose p-k}$, so
      $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
      Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
      If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We have ${p choose k} = {p choose p-k}$, so
        $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
        Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
        If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.






        share|cite|improve this answer









        $endgroup$



        We have ${p choose k} = {p choose p-k}$, so
        $$ sum_{k=0}^{lfloor p/2 rfloor} {p choose k} = sum_{k = p - lfloor p/2 rfloor}^p {p choose k}$$
        Thus if $p$ is odd, ${0, ldots, lfloor p/2 rfloor }$ and ${p - lfloor p/2 rfloor, ldots, p}$ are disjoint and their union is ${0, ldots, p}$, so using the binomial theorem your sum is $2^{p-1}$.
        If $p$ is even, $p/2$ is in both sets, so your sum is $2^{p-1} + {p choose p/2}/2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 13 at 21:13









        Robert IsraelRobert Israel

        328k23216469




        328k23216469






























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