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











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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.






summation binomial-coefficients floor-function







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    Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
    – Robert Z
    Nov 17 at 15:57










  • 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.
    – Joel Pereira
    Nov 17 at 16:02















up vote
0
down vote

favorite












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.






summation binomial-coefficients floor-function







share|cite|improve this question


















  • 1




    Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
    – Robert Z
    Nov 17 at 15:57










  • 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.
    – Joel Pereira
    Nov 17 at 16:02













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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.






summation binomial-coefficients floor-function







share|cite|improve this question













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.






summation binomial-coefficients floor-function




summation binomial-coefficients floor-function






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asked Nov 17 at 15:54









FuzzyPixelz

337214




337214








  • 1




    Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
    – Robert Z
    Nov 17 at 15:57










  • 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.
    – Joel Pereira
    Nov 17 at 16:02














  • 1




    Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
    – Robert Z
    Nov 17 at 15:57










  • 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.
    – Joel Pereira
    Nov 17 at 16:02








1




1




Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57




Recall that $sum_{k=0}^pbinom{p}{k}=(1+1)^k$.
– Robert Z
Nov 17 at 15:57












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.
– Joel Pereira
Nov 17 at 16:02




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.
– Joel Pereira
Nov 17 at 16:02










1 Answer
1






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0
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You can write
$$
eqalign{
& 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
p cr
p - k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
p cr
k cr} right)} = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) = cr
& = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
p cr
k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
p cr
leftlfloor {{p over 2}} rightrfloor cr} right) cr}
$$






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    1 Answer
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    up vote
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    You can write
    $$
    eqalign{
    & 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
    p cr
    k cr} right)} = cr
    & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
    p cr
    k cr} right)} = cr
    & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
    p cr
    p - k cr} right)} = cr
    & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
    p cr
    k cr} right)} = cr
    & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
    p cr
    k cr} right)} = cr
    & = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
    p cr
    k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
    p cr
    k cr} right)} = cr
    & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
    p cr
    leftlfloor {{p over 2}} rightrfloor cr} right) = cr
    & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
    p cr
    k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
    p cr
    leftlfloor {{p over 2}} rightrfloor cr} right) cr}
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      You can write
      $$
      eqalign{
      & 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
      p cr
      k cr} right)} = cr
      & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
      p cr
      k cr} right)} = cr
      & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
      p cr
      p - k cr} right)} = cr
      & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
      p cr
      k cr} right)} = cr
      & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
      p cr
      k cr} right)} = cr
      & = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
      p cr
      k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
      p cr
      k cr} right)} = cr
      & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
      p cr
      leftlfloor {{p over 2}} rightrfloor cr} right) = cr
      & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
      p cr
      k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
      p cr
      leftlfloor {{p over 2}} rightrfloor cr} right) cr}
      $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        You can write
        $$
        eqalign{
        & 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
        p cr
        p - k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
        p cr
        k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
        p cr
        k cr} right)} = cr
        & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
        p cr
        leftlfloor {{p over 2}} rightrfloor cr} right) = cr
        & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
        p cr
        leftlfloor {{p over 2}} rightrfloor cr} right) cr}
        $$






        share|cite|improve this answer












        You can write
        $$
        eqalign{
        & 2^{,p} = sumlimits_{0, le ,k, le ,p} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{leftlfloor {{p over 2}} rightrfloor + 1, le ,k, le ,p} {left( matrix{
        p cr
        p - k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{0, le ,k, le ,p - leftlfloor {{p over 2}} rightrfloor - 1} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
        p cr
        k cr} right)} = cr
        & = sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
        p cr
        k cr} right)} + sumlimits_{leftlceil {{p over 2}} rightrceil , le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + sumlimits_{0, le ,k, le ,leftlceil {{p over 2}} rightrceil - 1} {left( matrix{
        p cr
        k cr} right)} = cr
        & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} + left( {leftlceil {{p over 2}} rightrceil - leftlfloor {{p over 2}} rightrfloor - 1} right)left( matrix{
        p cr
        leftlfloor {{p over 2}} rightrfloor cr} right) = cr
        & = 2sumlimits_{0, le ,k, le ,leftlfloor {{p over 2}} rightrfloor } {left( matrix{
        p cr
        k cr} right)} - left( {1 - pbmod 2} right)left( matrix{
        p cr
        leftlfloor {{p over 2}} rightrfloor cr} right) cr}
        $$







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        answered Nov 18 at 12:13









        G Cab

        17.1k31237




        17.1k31237






























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