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combinatorics floor summation

Proof that $sum_{k=0}^m binom{m}{k}frac{1}{k+1} = frac{2^{m+1}-1}{m+1}$Summation involving binomial coefficients: $sum_{i=0}^{100} binom{300}{3i}$summation combinatoric again with floor functionProof of the binomial identity $displaystylebinom{m}{n}=sum_{k=0}^{lfloor n/2 rfloor} 2^{1-delta_{k,n-k}} binom{m/2}{k} binom{m/2}{n-k}$Proof/derivation of $limlimits_{ntoinfty}{frac1{2^n}sumlimits_{k=0}^nbinom{n}{k}frac{an+bk}{cn+dk}}stackrel?=frac{2a+b}{2c+d}$?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 $sum_{k=0}^{n}4^k binom{n}{k}$

2

$begingroup$

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

$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$

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
  53 mins 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.

$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$

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
  53 mins 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.

$sum_{i=0}^{floor(frac{n}{2})} binom{n}{2i}p^{2i}(1-p)^{n-2i} = frac{1}{2}((2p-1)^{n}+1)$

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:

begin{array}{rrl}
&(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
+&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
&(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
end{array}
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 49 mins ago

answered 56 mins ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that's a very neat proof!
  $endgroup$
  – SzymonSzymon
  49 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
$$
begin{align}
&frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
&= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
&= frac12((1-2p)^n + 1),
end{align}
$$
as required.

share|cite|improve this answer

answered 38 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:

begin{array}{rrl}
&(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
+&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
&(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
end{array}
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 49 mins ago

answered 56 mins ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

2

$begingroup$

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

begin{array}{rrl}
&(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
+&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
&(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
end{array}
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 49 mins ago

answered 56 mins ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

begin{array}{rrl}
&(p+q)^n&=;;sum_{i=0}^n binom{n}ip^iq^{n-i}\
+&(-p+q)^n&=;;sum_{i=0}^n binom{n}i(-p)^iq^{n-i}\hline
&(p+q)^n+(-p+q)^n &=2sum_{i=0}^{lfloor n/2rfloor}binom{n}{2i}p^{2i}q^{n-2i}
end{array}
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 49 mins ago

answered 56 mins ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

share|cite|improve this answer

edited 49 mins ago

answered 56 mins ago

Mike EarnestMike Earnest

24.8k22151

answered 56 mins ago

Mike EarnestMike Earnest

24.8k22151

answered 56 mins 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
$$
begin{align}
&frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
&= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
&= frac12((1-2p)^n + 1),
end{align}
$$
as required.

share|cite|improve this answer

answered 38 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
$$
begin{align}
&frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
&= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
&= frac12((1-2p)^n + 1),
end{align}
$$
as required.

share|cite|improve this answer

answered 38 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
$$
begin{align}
&frac12((1-2p)^{n-1}+1)(1-p) + (1 - frac12((1-2p)^{n-1}+1))cdot p\
&= frac12(1-2p)^{n-1}(1-p-p) +frac12(1-p-p) + p\
&= frac12((1-2p)^n + 1),
end{align}
$$
as required.

share|cite|improve this answer

answered 38 mins ago

FredHFredH

2,144914

$endgroup$

share|cite|improve this answer

answered 38 mins ago

FredHFredH

2,144914

answered 38 mins ago

FredHFredH

2,144914

answered 38 mins ago

FredHFredH

2,144914