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How can I prove algebraically the sums of binomial coefficients formula? [duplicate]

Sum of the rows of Pascal's Triangle.Proving by induction that $ sum_k=0^nn choose k = 2^n$Alternative proof that $sum_j=0^nbinomnj = binomn0 + cdots + binomnn = 2^n$Two inequalities with binomial coefficientsProve summation formula for binomial coefficientsBinomial coefficients as multiple sumsCombinatorial proof with binomial coefficientsHow to prove this algebraically?Identities involving binomial coefficients, floors, and ceilingsBinomial coefficients identityCombinatorial Proof, Binomial CoefficientsHow to prove a modified version of the binomial theorem?Combinatorial proofs of several binomial sums

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This question already has an answer here:

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  Sum of the rows of Pascal's Triangle.
  
  3 answers
  
  

I want to know a simple way to prove $$sum_k=0^n nchoosek = 2^n.$$ Our teacher only gave us the combinatorial proof.

combinatorics

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edited Mar 29 at 1:59

gt6989b

35.5k22557

asked Mar 29 at 1:53

ydnfmewydnfmew

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marked as duplicate by Mike Earnest, José Carlos Santos, max_zorn, dantopa, Adrian Keister Mar 29 at 21:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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  This question on cross validated.
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  – Brian
  Mar 29 at 1:56
  

  
  
  
  


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  IMHO, the combinatorial way IS the simple way. The other fun one is to expand $(x+y)^n$ and toss in $x=y=1$.
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  – Randall
  Mar 29 at 1:56
  

  
  
  
  


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  Personally, I don't feel the combinatorial way is necessarily the "simpler" way, depending on your exposure. Though I will agree it's at least the more insightful and interesting way.
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  – Eevee Trainer
  Mar 29 at 2:00
  

  
  
  
  


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  If I were sitting in on your class, I would shake my finger at your teacher for not showing you the easy proof, as given by @EeveeTrainer below.
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  – Lubin
  Mar 29 at 2:04
  

  
  
  
  


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  @Lubin I might disagree. If you need to prove the BT first, it's much more annoying and gross.
  $endgroup$
  – Randall
  Mar 29 at 2:06
  

  
  
  
  

 | 
show 4 more comments

2

$begingroup$

This question already has an answer here:

• 
  Sum of the rows of Pascal's Triangle.
  
  3 answers
  
  

I want to know a simple way to prove $$sum_k=0^n nchoosek = 2^n.$$ Our teacher only gave us the combinatorial proof.

combinatorics

share|cite|improve this question

edited Mar 29 at 1:59

gt6989b

35.5k22557

asked Mar 29 at 1:53

ydnfmewydnfmew

261

New contributor

ydnfmew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

marked as duplicate by Mike Earnest, José Carlos Santos, max_zorn, dantopa, Adrian Keister Mar 29 at 21:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This question on cross validated.
  $endgroup$
  – Brian
  Mar 29 at 1:56
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  IMHO, the combinatorial way IS the simple way. The other fun one is to expand $(x+y)^n$ and toss in $x=y=1$.
  $endgroup$
  – Randall
  Mar 29 at 1:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Personally, I don't feel the combinatorial way is necessarily the "simpler" way, depending on your exposure. Though I will agree it's at least the more insightful and interesting way.
  $endgroup$
  – Eevee Trainer
  Mar 29 at 2:00
  

  
  
  
  


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  $begingroup$
  If I were sitting in on your class, I would shake my finger at your teacher for not showing you the easy proof, as given by @EeveeTrainer below.
  $endgroup$
  – Lubin
  Mar 29 at 2:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lubin I might disagree. If you need to prove the BT first, it's much more annoying and gross.
  $endgroup$
  – Randall
  Mar 29 at 2:06
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

This question already has an answer here:

• 
  Sum of the rows of Pascal's Triangle.
  
  3 answers
  
  

I want to know a simple way to prove $$sum_k=0^n nchoosek = 2^n.$$ Our teacher only gave us the combinatorial proof.

combinatorics

share|cite|improve this question

edited Mar 29 at 1:59

gt6989b

35.5k22557

asked Mar 29 at 1:53

ydnfmewydnfmew

261

New contributor

ydnfmew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited Mar 29 at 1:59

gt6989b

35.5k22557

asked Mar 29 at 1:53

ydnfmewydnfmew

261

New contributor

ydnfmew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

edited Mar 29 at 1:59

gt6989b

35.5k22557

asked Mar 29 at 1:53

ydnfmewydnfmew

261

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edited Mar 29 at 1:59

gt6989b

35.5k22557

edited Mar 29 at 1:59

gt6989b

35.5k22557

asked Mar 29 at 1:53

ydnfmewydnfmew

261

New contributor

ydnfmew is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Mar 29 at 1:53

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1 Answer
1

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4

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Let's assume the binomial theorem can be taken for granted. By it, for $n in Bbb N$,

$$(a+b)^n = sum_k=0^n binom n k a^k b^n-k$$

Take $a=b=1$. Then we have

$$(1+1)^n = colorblue 2^n = sum_k=0^n binom n k = sum_k=0^n binom n k 1^k 1^n-k$$

since $1$ to any power is just $1$.

share|cite|improve this answer

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740

$endgroup$

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  $begingroup$
  Just to note, the simplest proof of the binomial theorem (admittedly subjective, as you point out in your comment) for positive integral powers is probably the combinatorial one: The coefficient of $a^kb^n - k$ in the expansion of $(a + b)^n$ is the number of ways of choosing $k$ $a$'s and $n - k$ $b$'s from the $n$ brackets $(a + b)$, which is the number of ways of choosing $k$ brackets out of $n$, from which the $a$'s can be chosen, which is $binom n k$.
  $endgroup$
  – M. Vinay
  Mar 29 at 2:49
  

  
  
  
  


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  Gosh, @M.Vinay, I would define $binom nk$ recursively, right outta the Pascal triangle. The proof drops out that way too.
  $endgroup$
  – Lubin
  Mar 29 at 4:32
  

  
  
  
  

add a comment | 

4

$begingroup$

Let's assume the binomial theorem can be taken for granted. By it, for $n in Bbb N$,

$$(a+b)^n = sum_k=0^n binom n k a^k b^n-k$$

Take $a=b=1$. Then we have

$$(1+1)^n = colorblue 2^n = sum_k=0^n binom n k = sum_k=0^n binom n k 1^k 1^n-k$$

since $1$ to any power is just $1$.

share|cite|improve this answer

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just to note, the simplest proof of the binomial theorem (admittedly subjective, as you point out in your comment) for positive integral powers is probably the combinatorial one: The coefficient of $a^kb^n - k$ in the expansion of $(a + b)^n$ is the number of ways of choosing $k$ $a$'s and $n - k$ $b$'s from the $n$ brackets $(a + b)$, which is the number of ways of choosing $k$ brackets out of $n$, from which the $a$'s can be chosen, which is $binom n k$.
  $endgroup$
  – M. Vinay
  Mar 29 at 2:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Gosh, @M.Vinay, I would define $binom nk$ recursively, right outta the Pascal triangle. The proof drops out that way too.
  $endgroup$
  – Lubin
  Mar 29 at 4:32
  

  
  
  
  

add a comment | 

4

$begingroup$

Let's assume the binomial theorem can be taken for granted. By it, for $n in Bbb N$,

$$(a+b)^n = sum_k=0^n binom n k a^k b^n-k$$

Take $a=b=1$. Then we have

$$(1+1)^n = colorblue 2^n = sum_k=0^n binom n k = sum_k=0^n binom n k 1^k 1^n-k$$

since $1$ to any power is just $1$.

share|cite|improve this answer

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just to note, the simplest proof of the binomial theorem (admittedly subjective, as you point out in your comment) for positive integral powers is probably the combinatorial one: The coefficient of $a^kb^n - k$ in the expansion of $(a + b)^n$ is the number of ways of choosing $k$ $a$'s and $n - k$ $b$'s from the $n$ brackets $(a + b)$, which is the number of ways of choosing $k$ brackets out of $n$, from which the $a$'s can be chosen, which is $binom n k$.
  $endgroup$
  – M. Vinay
  Mar 29 at 2:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Gosh, @M.Vinay, I would define $binom nk$ recursively, right outta the Pascal triangle. The proof drops out that way too.
  $endgroup$
  – Lubin
  Mar 29 at 4:32
  

  
  
  
  

add a comment | 

$begingroup$

Let's assume the binomial theorem can be taken for granted. By it, for $n in Bbb N$,

$$(a+b)^n = sum_k=0^n binom n k a^k b^n-k$$

Take $a=b=1$. Then we have

$$(1+1)^n = colorblue 2^n = sum_k=0^n binom n k = sum_k=0^n binom n k 1^k 1^n-k$$

since $1$ to any power is just $1$.

share|cite|improve this answer

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740

$endgroup$

share|cite|improve this answer

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740

answered Mar 29 at 1:59

Eevee TrainerEevee Trainer

9,67531740