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the coefficient of $ x^n-1$ in $(x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots(x+2)^n$

1

$begingroup$

Find the coefficient of $ x^n-1$ in:

$ \ (x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots+(x+2)^n$

My Approach:

$(x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots+(x+2)^n=sum(x+3)^n-r(x+2)^r$

For each term of this summation:
we have to calculate the number of ways of choosing one (x+a) for which we would multiply the 'a' and not the 'x' part. They can be chosen in $binomn-r1+binomr1$ ways. In case the chosen is (x-3) the coefficient of $x^n-1$ would be 3 else it would be 2. Hence the coefficient of $x^n-1$ for the general term would be $3binomn-r1+2binomr1$.

This simplifies to $3(n-r)+2r=3n-r$. The coefficient of $x^n-1$ in the summation is $sum_0^n 3n-sum_r=0^nn r=3n^2-fracn(n+1)2=frac6n^2-n^2-n2=frac5n^2-n2$.

The answer given is $5binomn+12$ but I can't figure out what the error is in this logic. It would be great if someone could help me answer this question.

binomial-coefficients combinations binomial-theorem

share|cite|improve this question

edited Feb 20 '17 at 11:44

Osheen Sachdev

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

$endgroup$

add a comment | 

1

$begingroup$

Find the coefficient of $ x^n-1$ in:

$ \ (x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots+(x+2)^n$

My Approach:

$(x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots+(x+2)^n=sum(x+3)^n-r(x+2)^r$

For each term of this summation:
we have to calculate the number of ways of choosing one (x+a) for which we would multiply the 'a' and not the 'x' part. They can be chosen in $binomn-r1+binomr1$ ways. In case the chosen is (x-3) the coefficient of $x^n-1$ would be 3 else it would be 2. Hence the coefficient of $x^n-1$ for the general term would be $3binomn-r1+2binomr1$.

This simplifies to $3(n-r)+2r=3n-r$. The coefficient of $x^n-1$ in the summation is $sum_0^n 3n-sum_r=0^nn r=3n^2-fracn(n+1)2=frac6n^2-n^2-n2=frac5n^2-n2$.

The answer given is $5binomn+12$ but I can't figure out what the error is in this logic. It would be great if someone could help me answer this question.

binomial-coefficients combinations binomial-theorem

share|cite|improve this question

edited Feb 20 '17 at 11:44

Osheen Sachdev

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

$endgroup$

add a comment | 

$begingroup$

Find the coefficient of $ x^n-1$ in:

$ \ (x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots+(x+2)^n$

My Approach:

$(x+3)^n+(x+3)^n-1(x+2)+(x+3)^n-2(x+2)^2cdots+(x+2)^n=sum(x+3)^n-r(x+2)^r$

For each term of this summation:
we have to calculate the number of ways of choosing one (x+a) for which we would multiply the 'a' and not the 'x' part. They can be chosen in $binomn-r1+binomr1$ ways. In case the chosen is (x-3) the coefficient of $x^n-1$ would be 3 else it would be 2. Hence the coefficient of $x^n-1$ for the general term would be $3binomn-r1+2binomr1$.

This simplifies to $3(n-r)+2r=3n-r$. The coefficient of $x^n-1$ in the summation is $sum_0^n 3n-sum_r=0^nn r=3n^2-fracn(n+1)2=frac6n^2-n^2-n2=frac5n^2-n2$.

The answer given is $5binomn+12$ but I can't figure out what the error is in this logic. It would be great if someone could help me answer this question.

binomial-coefficients combinations binomial-theorem

share|cite|improve this question

edited Feb 20 '17 at 11:44

Osheen Sachdev

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

$endgroup$

share|cite|improve this question

edited Feb 20 '17 at 11:44

Osheen Sachdev

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

share|cite|improve this question

edited Feb 20 '17 at 11:44

Osheen Sachdev

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

asked Feb 20 '17 at 11:32

Osheen SachdevOsheen Sachdev

1,23021641

1 Answer
1

active

oldest

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3

$begingroup$

It is easier to simplify the given expression by using formula $$a^r-1+a^r-2b+cdots+ab^r-2 +b^r-1=fraca^r-b^ra-b$$ which is easily proved by multiplying both sides of the equation by $a-b$. Here we have $r=n+1,a=x+3,b=x+2$ and hence the expression simplifies to $(x+3)^n+1-(x+2)^n+1$ and the desired coefficient is $$binomn+12(3^2-2^2)=5binomn+12$$

---

You had your approach correct but you made a calculation mistake while calculating the sum $sum_r=0^n3n$ which should evaluate to $3n(n+1)$ but you wrote $3n^2$.

share|cite|improve this answer

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is my way :)
  $endgroup$
  – lab bhattacharjee
  Feb 20 '17 at 12:22
  

  
  
  
  

add a comment | 

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3

$begingroup$

It is easier to simplify the given expression by using formula $$a^r-1+a^r-2b+cdots+ab^r-2 +b^r-1=fraca^r-b^ra-b$$ which is easily proved by multiplying both sides of the equation by $a-b$. Here we have $r=n+1,a=x+3,b=x+2$ and hence the expression simplifies to $(x+3)^n+1-(x+2)^n+1$ and the desired coefficient is $$binomn+12(3^2-2^2)=5binomn+12$$

---

You had your approach correct but you made a calculation mistake while calculating the sum $sum_r=0^n3n$ which should evaluate to $3n(n+1)$ but you wrote $3n^2$.

share|cite|improve this answer

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is my way :)
  $endgroup$
  – lab bhattacharjee
  Feb 20 '17 at 12:22
  

  
  
  
  

add a comment | 

3

$begingroup$

It is easier to simplify the given expression by using formula $$a^r-1+a^r-2b+cdots+ab^r-2 +b^r-1=fraca^r-b^ra-b$$ which is easily proved by multiplying both sides of the equation by $a-b$. Here we have $r=n+1,a=x+3,b=x+2$ and hence the expression simplifies to $(x+3)^n+1-(x+2)^n+1$ and the desired coefficient is $$binomn+12(3^2-2^2)=5binomn+12$$

---

You had your approach correct but you made a calculation mistake while calculating the sum $sum_r=0^n3n$ which should evaluate to $3n(n+1)$ but you wrote $3n^2$.

share|cite|improve this answer

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is my way :)
  $endgroup$
  – lab bhattacharjee
  Feb 20 '17 at 12:22
  

  
  
  
  

add a comment | 

$begingroup$

It is easier to simplify the given expression by using formula $$a^r-1+a^r-2b+cdots+ab^r-2 +b^r-1=fraca^r-b^ra-b$$ which is easily proved by multiplying both sides of the equation by $a-b$. Here we have $r=n+1,a=x+3,b=x+2$ and hence the expression simplifies to $(x+3)^n+1-(x+2)^n+1$ and the desired coefficient is $$binomn+12(3^2-2^2)=5binomn+12$$

---

You had your approach correct but you made a calculation mistake while calculating the sum $sum_r=0^n3n$ which should evaluate to $3n(n+1)$ but you wrote $3n^2$.

share|cite|improve this answer

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170

$endgroup$

share|cite|improve this answer

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170

answered Feb 20 '17 at 11:59

Paramanand SinghParamanand Singh

51.4k560170