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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?

probability combinatorics summation binomial-coefficients binomial-distribution

share|cite|improve this question

edited 1 hour ago

Austin Mohr

20.5k35098

asked 1 hour ago

SzymonSzymonSzymonSzymon

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.

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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?

probability combinatorics summation binomial-coefficients binomial-distribution

share|cite|improve this question

edited 1 hour ago

Austin Mohr

20.5k35098

asked 1 hour ago

SzymonSzymonSzymonSzymon

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.

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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?

probability combinatorics summation binomial-coefficients binomial-distribution

share|cite|improve this question

edited 1 hour ago

Austin Mohr

20.5k35098

asked 1 hour ago

SzymonSzymonSzymonSzymon

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.

$endgroup$

share|cite|improve this question

edited 1 hour ago

Austin Mohr

20.5k35098

asked 1 hour ago

SzymonSzymonSzymonSzymon

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.

share|cite|improve this question

edited 1 hour ago

Austin Mohr

20.5k35098

asked 1 hour ago

SzymonSzymonSzymonSzymon

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.

edited 1 hour ago

Austin Mohr

20.5k35098

edited 1 hour ago

Austin Mohr

20.5k35098

asked 1 hour ago

SzymonSzymonSzymonSzymon

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.

asked 1 hour ago

SzymonSzymonSzymonSzymon

111

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

edited 57 mins ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

answered 47 mins ago

FredHFredH

2,144914

$endgroup$

add a comment | 

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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

edited 57 mins ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

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

edited 57 mins ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 57 mins ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

share|cite|improve this answer

edited 57 mins ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

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

answered 47 mins ago

FredHFredH

2,144914

$endgroup$

add a comment | 

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

answered 47 mins ago

FredHFredH

2,144914

$endgroup$

add a comment | 

$begingroup$

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

answered 47 mins ago

FredHFredH

2,144914

$endgroup$

share|cite|improve this answer

answered 47 mins ago

FredHFredH

2,144914

answered 47 mins ago

FredHFredH

2,144914

answered 47 mins ago

FredHFredH

2,144914