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Find the coefficient of $x^n.$ [closed]

-2

1

$begingroup$

How do you recommend that I find the coefficient of $x^n$ given the expansion of

$$(1+x)^2n+ x(1+x)^2n-1 + x^2(1+x)^2n-2+ dots + x^n(1+x)^n?$$

I already tried a few different ways but I can't get it.

binomial-coefficients

share|cite|improve this question

edited Mar 31 at 5:45

Dbchatto67

2,787622

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

$endgroup$

closed as off-topic by Shailesh, Paul Frost, Javi, Saad, Thomas Shelby Mar 31 at 11:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Paul Frost, Javi, Saad, Thomas Shelby

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks Eevee already arrenged it
  $endgroup$
  – Rúdi Gualter de Oliveira
  Mar 31 at 5:29
  

  
  
  
  

add a comment | 

-2

1

$begingroup$

How do you recommend that I find the coefficient of $x^n$ given the expansion of

$$(1+x)^2n+ x(1+x)^2n-1 + x^2(1+x)^2n-2+ dots + x^n(1+x)^n?$$

I already tried a few different ways but I can't get it.

binomial-coefficients

share|cite|improve this question

edited Mar 31 at 5:45

Dbchatto67

2,787622

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

$endgroup$

closed as off-topic by Shailesh, Paul Frost, Javi, Saad, Thomas Shelby Mar 31 at 11:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Shailesh, Paul Frost, Javi, Saad, Thomas Shelby

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks Eevee already arrenged it
  $endgroup$
  – Rúdi Gualter de Oliveira
  Mar 31 at 5:29
  

  
  
  
  

add a comment | 

$begingroup$

How do you recommend that I find the coefficient of $x^n$ given the expansion of

$$(1+x)^2n+ x(1+x)^2n-1 + x^2(1+x)^2n-2+ dots + x^n(1+x)^n?$$

I already tried a few different ways but I can't get it.

binomial-coefficients

share|cite|improve this question

edited Mar 31 at 5:45

Dbchatto67

2,787622

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

$endgroup$

share|cite|improve this question

edited Mar 31 at 5:45

Dbchatto67

2,787622

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

share|cite|improve this question

edited Mar 31 at 5:45

Dbchatto67

2,787622

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

edited Mar 31 at 5:45

Dbchatto67

2,787622

edited Mar 31 at 5:45

Dbchatto67

2,787622

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

asked Mar 31 at 5:17

Rúdi Gualter de OliveiraRúdi Gualter de Oliveira

62

2 Answers
2

active

oldest

votes

3

$begingroup$

You have
$$S(x)=sum_k=0^n x^k(1+x)^2n-k.$$
This is a GP, so
$$S(x)=frac(1+x)^2n+1-x^n+1(1+x)^n1+x-x=(1+x)^2n+1-x^n+1(1+x)^n.$$
The coefficient of $x^n$ in $x^n+1(1+x)^n$ is zero, so the coefficient of
$x^n$ in $S(x)$ is that of $x^n$ in $(1+x)^2n+1$, that is $binom2n+1n$.

share|cite|improve this answer

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks I see, thanks very much
  $endgroup$
  – Rúdi Gualter de Oliveira
  Mar 31 at 5:40
  

  
  
  
  

add a comment | 

0

$begingroup$

The coefficient of $x^n$ is $binom 2n n + binom 2n-1 n-1 + binom 2n-2 n-2 + cdots + binom n 0 = binom 2n+1 n.$

share|cite|improve this answer

edited Mar 31 at 5:41

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Who has downvoted my answer without prior reason?
  $endgroup$
  – Dbchatto67
  Mar 31 at 5:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's that @Rudi?
  $endgroup$
  – Dbchatto67
  Mar 31 at 5:40
  

  
  
  
  

add a comment | 

3

$begingroup$

You have
$$S(x)=sum_k=0^n x^k(1+x)^2n-k.$$
This is a GP, so
$$S(x)=frac(1+x)^2n+1-x^n+1(1+x)^n1+x-x=(1+x)^2n+1-x^n+1(1+x)^n.$$
The coefficient of $x^n$ in $x^n+1(1+x)^n$ is zero, so the coefficient of
$x^n$ in $S(x)$ is that of $x^n$ in $(1+x)^2n+1$, that is $binom2n+1n$.

share|cite|improve this answer

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks I see, thanks very much
  $endgroup$
  – Rúdi Gualter de Oliveira
  Mar 31 at 5:40
  

  
  
  
  

add a comment | 

3

$begingroup$

You have
$$S(x)=sum_k=0^n x^k(1+x)^2n-k.$$
This is a GP, so
$$S(x)=frac(1+x)^2n+1-x^n+1(1+x)^n1+x-x=(1+x)^2n+1-x^n+1(1+x)^n.$$
The coefficient of $x^n$ in $x^n+1(1+x)^n$ is zero, so the coefficient of
$x^n$ in $S(x)$ is that of $x^n$ in $(1+x)^2n+1$, that is $binom2n+1n$.

share|cite|improve this answer

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks I see, thanks very much
  $endgroup$
  – Rúdi Gualter de Oliveira
  Mar 31 at 5:40
  

  
  
  
  

add a comment | 

$begingroup$

You have
$$S(x)=sum_k=0^n x^k(1+x)^2n-k.$$
This is a GP, so
$$S(x)=frac(1+x)^2n+1-x^n+1(1+x)^n1+x-x=(1+x)^2n+1-x^n+1(1+x)^n.$$
The coefficient of $x^n$ in $x^n+1(1+x)^n$ is zero, so the coefficient of
$x^n$ in $S(x)$ is that of $x^n$ in $(1+x)^2n+1$, that is $binom2n+1n$.

share|cite|improve this answer

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

$endgroup$

share|cite|improve this answer

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

answered Mar 31 at 5:37

Lord Shark the UnknownLord Shark the Unknown

108k1162136

0

$begingroup$

The coefficient of $x^n$ is $binom 2n n + binom 2n-1 n-1 + binom 2n-2 n-2 + cdots + binom n 0 = binom 2n+1 n.$

share|cite|improve this answer

edited Mar 31 at 5:41

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Who has downvoted my answer without prior reason?
  $endgroup$
  – Dbchatto67
  Mar 31 at 5:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's that @Rudi?
  $endgroup$
  – Dbchatto67
  Mar 31 at 5:40
  

  
  
  
  

add a comment | 

0

$begingroup$

The coefficient of $x^n$ is $binom 2n n + binom 2n-1 n-1 + binom 2n-2 n-2 + cdots + binom n 0 = binom 2n+1 n.$

share|cite|improve this answer

edited Mar 31 at 5:41

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Who has downvoted my answer without prior reason?
  $endgroup$
  – Dbchatto67
  Mar 31 at 5:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's that @Rudi?
  $endgroup$
  – Dbchatto67
  Mar 31 at 5:40
  

  
  
  
  

add a comment | 

$begingroup$

The coefficient of $x^n$ is $binom 2n n + binom 2n-1 n-1 + binom 2n-2 n-2 + cdots + binom n 0 = binom 2n+1 n.$

share|cite|improve this answer

edited Mar 31 at 5:41

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622

$endgroup$

share|cite|improve this answer

edited Mar 31 at 5:41

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622

answered Mar 31 at 5:31

Dbchatto67Dbchatto67

2,787622