combinatorics floor summationProof that $sum_k=0^m binommkfrac1k+1 = frac2^m+1-1m+1$Summation involving binomial coefficients: $sum_i=0^100 binom3003i$summation combinatoric again with floor functionProof of the binomial identity $displaystylebinommn=sum_k=0^lfloor n/2 rfloor 2^1-delta_k,n-k binomm/2k binomm/2n-k$Proof/derivation of $limlimits_ntoinftyfrac12^nsumlimits_k=0^nbinomnkfracan+bkcn+dkstackrel?=frac2a+b2c+d$?How to find documentation for or a proof of the following known binomial identityEvaluate summation combinatoricsWeird Combinatorial IdentitiyProve the following by two different methods, one combinatorial and one algebraicFind $sum_k=0^n4^k binomnk$

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combinatorics floor summation


Proof that $sum_k=0^m binommkfrac1k+1 = frac2^m+1-1m+1$Summation involving binomial coefficients: $sum_i=0^100 binom3003i$summation combinatoric again with floor functionProof of the binomial identity $displaystylebinommn=sum_k=0^lfloor n/2 rfloor 2^1-delta_k,n-k binomm/2k binomm/2n-k$Proof/derivation of $limlimits_ntoinftyfrac12^nsumlimits_k=0^nbinomnkfracan+bkcn+dkstackrel?=frac2a+b2c+d$?How to find documentation for or a proof of the following known binomial identityEvaluate summation combinatoricsWeird Combinatorial IdentitiyProve the following by two different methods, one combinatorial and one algebraicFind $sum_k=0^n4^k binomnk$













2












$begingroup$


While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_i=0^floor(fracn2) binomn2ip^2i(1-p)^n-2i = frac12((2p-1)^n+1)$



Any ideas?










share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    1 hour ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    1 hour ago















2












$begingroup$


While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_i=0^floor(fracn2) binomn2ip^2i(1-p)^n-2i = frac12((2p-1)^n+1)$



Any ideas?










share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    1 hour ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    1 hour ago













2












2








2





$begingroup$


While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_i=0^floor(fracn2) binomn2ip^2i(1-p)^n-2i = frac12((2p-1)^n+1)$



Any ideas?










share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




While solving a problem I came across a rather interesting identity, and I do not see how I could prove it.



$sum_i=0^floor(fracn2) binomn2ip^2i(1-p)^n-2i = frac12((2p-1)^n+1)$



Any ideas?







probability combinatorics summation binomial-coefficients binomial-distribution






share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Austin Mohr

20.5k35098




20.5k35098






New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago









SzymonSzymonSzymonSzymon

111




111




New contributor




SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






SzymonSzymon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    1 hour ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    1 hour ago












  • 1




    $begingroup$
    Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
    $endgroup$
    – Austin Mohr
    1 hour ago






  • 1




    $begingroup$
    I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
    $endgroup$
    – Mike Earnest
    1 hour ago







1




1




$begingroup$
Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
$endgroup$
– Austin Mohr
1 hour ago




$begingroup$
Flip $n$ coins each having probability $p$ of landing heads. The lefthand side is the probability of flipping an even number of heads. I don't immediately see how the righthand side calculates the same probability.
$endgroup$
– Austin Mohr
1 hour ago




1




1




$begingroup$
I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
$endgroup$
– Mike Earnest
1 hour ago




$begingroup$
I believe you have a mistake: the RHS should be $frac12((1-2p)^n+1)$.
$endgroup$
– Mike Earnest
1 hour ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $q=1-p$. Apply the binomial theorem twice, then add:



beginarrayrrl
&(p+q)^n&=;;sum_i=0^n binomnip^iq^n-i\
+&(-p+q)^n&=;;sum_i=0^n binomni(-p)^iq^n-i\hline
&(p+q)^n+(-p+q)^n &=2sum_i=0^lfloor n/2rfloorbinomn2ip^2iq^n-2i
endarray

To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Thank you, that's a very neat proof!
    $endgroup$
    – SzymonSzymon
    1 hour ago


















1












$begingroup$

With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
$$
beginalign
&frac12((1-2p)^n-1+1)(1-p) + (1 - frac12((1-2p)^n-1+1))cdot p\
&= frac12(1-2p)^n-1(1-p-p) +frac12(1-p-p) + p\
&= frac12((1-2p)^n + 1),
endalign
$$

as required.






share|cite|improve this answer









$endgroup$












    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Let $q=1-p$. Apply the binomial theorem twice, then add:



    beginarrayrrl
    &(p+q)^n&=;;sum_i=0^n binomnip^iq^n-i\
    +&(-p+q)^n&=;;sum_i=0^n binomni(-p)^iq^n-i\hline
    &(p+q)^n+(-p+q)^n &=2sum_i=0^lfloor n/2rfloorbinomn2ip^2iq^n-2i
    endarray

    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      1 hour ago















    2












    $begingroup$

    Let $q=1-p$. Apply the binomial theorem twice, then add:



    beginarrayrrl
    &(p+q)^n&=;;sum_i=0^n binomnip^iq^n-i\
    +&(-p+q)^n&=;;sum_i=0^n binomni(-p)^iq^n-i\hline
    &(p+q)^n+(-p+q)^n &=2sum_i=0^lfloor n/2rfloorbinomn2ip^2iq^n-2i
    endarray

    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      1 hour ago













    2












    2








    2





    $begingroup$

    Let $q=1-p$. Apply the binomial theorem twice, then add:



    beginarrayrrl
    &(p+q)^n&=;;sum_i=0^n binomnip^iq^n-i\
    +&(-p+q)^n&=;;sum_i=0^n binomni(-p)^iq^n-i\hline
    &(p+q)^n+(-p+q)^n &=2sum_i=0^lfloor n/2rfloorbinomn2ip^2iq^n-2i
    endarray

    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.






    share|cite|improve this answer











    $endgroup$



    Let $q=1-p$. Apply the binomial theorem twice, then add:



    beginarrayrrl
    &(p+q)^n&=;;sum_i=0^n binomnip^iq^n-i\
    +&(-p+q)^n&=;;sum_i=0^n binomni(-p)^iq^n-i\hline
    &(p+q)^n+(-p+q)^n &=2sum_i=0^lfloor n/2rfloorbinomn2ip^2iq^n-2i
    endarray

    To understand the RHS of the sum, note when $i$ is odd we have $p^i+(-p)^i=0$, while when $i$ is even, $p^i+(-p)^i=2p^i$. Replacing $q$ with $1-p$ makes the LHS $1+(1-2p)^n$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    Mike EarnestMike Earnest

    24.8k22151




    24.8k22151











    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      1 hour ago
















    • $begingroup$
      Thank you, that's a very neat proof!
      $endgroup$
      – SzymonSzymon
      1 hour ago















    $begingroup$
    Thank you, that's a very neat proof!
    $endgroup$
    – SzymonSzymon
    1 hour ago




    $begingroup$
    Thank you, that's a very neat proof!
    $endgroup$
    – SzymonSzymon
    1 hour ago











    1












    $begingroup$

    With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



    Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



    Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
    $$
    beginalign
    &frac12((1-2p)^n-1+1)(1-p) + (1 - frac12((1-2p)^n-1+1))cdot p\
    &= frac12(1-2p)^n-1(1-p-p) +frac12(1-p-p) + p\
    &= frac12((1-2p)^n + 1),
    endalign
    $$

    as required.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



      Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



      Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
      $$
      beginalign
      &frac12((1-2p)^n-1+1)(1-p) + (1 - frac12((1-2p)^n-1+1))cdot p\
      &= frac12(1-2p)^n-1(1-p-p) +frac12(1-p-p) + p\
      &= frac12((1-2p)^n + 1),
      endalign
      $$

      as required.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



        Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



        Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
        $$
        beginalign
        &frac12((1-2p)^n-1+1)(1-p) + (1 - frac12((1-2p)^n-1+1))cdot p\
        &= frac12(1-2p)^n-1(1-p-p) +frac12(1-p-p) + p\
        &= frac12((1-2p)^n + 1),
        endalign
        $$

        as required.






        share|cite|improve this answer









        $endgroup$



        With the corrected RHS of $frac12((1-2p)^n+1)$ and Austin Mohr's interpretation of the LHS as the probability of an even number of heads occurring in $n$ tosses of coin which comes up heads with probability $p$, the result is easily established by induction.



        Case $n = 0$: $frac12((1-2p)^0+1) = 1$, which is correct since zero tosses always produce zero heads.



        Case $n > 0$: An even number of heads after $n$ tosses occurs either by having an even number of heads in $n-1$ tosses followed by tails, or an odd number in $n-1$ tosses followed by heads. The probability of this is
        $$
        beginalign
        &frac12((1-2p)^n-1+1)(1-p) + (1 - frac12((1-2p)^n-1+1))cdot p\
        &= frac12(1-2p)^n-1(1-p-p) +frac12(1-p-p) + p\
        &= frac12((1-2p)^n + 1),
        endalign
        $$

        as required.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 51 mins ago









        FredHFredH

        2,144914




        2,144914




















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