Yhyjyk

Multi tool use

How to pronounce "I ♥ Huckabees"?

Life insurance that covers only simultaneous/dual deaths

Do the common programs (for example: "ls", "cat") in Linux and BSD come from the same source code?

My adviser wants to be the first author

The German vowel “a” changes to the English “i”

Converting a variable frequency to TTL HIGH and LOW levels, based on a fixed (possible non-fixed?) frequency

Are ETF trackers fundamentally better than individual stocks?

Is a party consisting of only a bard, a cleric, and a warlock functional long-term?

Bacteria contamination inside a thermos bottle

Why is the President allowed to veto a cancellation of emergency powers?

Official degrees of earth’s rotation per day

Why did it take so long to abandon sail after steamships were demonstrated?

Why does energy conservation give me the wrong answer in this inelastic collision problem?

How to make healing in an exploration game interesting

What exactly is this small puffer fish doing and how did it manage to accomplish such a feat?

How to terminate ping <dest> &

"of which" is correct here?

Why does overlay work only on the first tcolorbox?

Equivalents to the present tense

What did “the good wine” (τὸν καλὸν οἶνον) mean in John 2:10?

Knife as defense against stray dogs

Recruiter wants very extensive technical details about all of my previous work

I got the following comment from a reputed math journal. What does it mean?

How to write cleanly even if my character uses expletive language?

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 1 hour ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, that's a very neat proof!
  $endgroup$
  – SzymonSzymon
  1 hour 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 51 mins ago

FredHFredH

2,144914

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

SzymonSzymon is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3150868%2fcombinatorics-floor-summation%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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 1 hour ago

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

answered 1 hour ago

Mike EarnestMike Earnest

24.8k22151

$endgroup$

share|cite|improve this answer

edited 1 hour 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 51 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 51 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 51 mins ago

FredHFredH

2,144914

$endgroup$

share|cite|improve this answer

answered 51 mins ago

FredHFredH

2,144914

answered 51 mins ago

FredHFredH

2,144914

answered 51 mins ago

FredHFredH

2,144914