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Coefficient of $x^{n-3}$ in $prod^{n}_{k=1}(x-k)$

What is the coefficient of $x^{19}$ in the expansion of $ prod limits_{n=1}^{20} (x+n^2)$?Find coefficient of $x^8$ in $(1+x+x^2+x^3+…)^c$Computing coefficient of $x^n$Coeff. of $x^{97}$ in $f(x) = (x-1)cdot (x-2)cdot (x-3)cdot (x-4)cdot …(x-100)$Finding the coefficient of a term in an expansion.finding coefficients of $x^{47}$ in polynomial expressionCoefficient in a binomial expansion of geometric progressionproduct of terms taken $3$ at a time in polynomial expression$sqrt{n+sqrt[3]{n+sqrt[3]{n+cdots cdots cdots }}}$ is a natural number, ismaximum value of expression $(sqrt{-3+4x-x^2}+4)^2+(x-5)^2$

0

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$begingroup$

Finding Coefficient of $x^{n-3}$ in $displaystyle prod^{n}_{k=1}(x-k)$

what i try

$displaystyle prod^{n}_{k=1}(x-k)=x^n-bigg(sum^{n}_{i=1}ibigg)+bigg(mathop{sumsum}_{1leq i <j leq n}ijbigg)x^{n-2}-bigg(mathop{sumsumsum}_{1leq i <j<kleq n}ijkbigg)x^{n-3}+cdots $

How do i find $displaystyle bigg(mathop{sumsumsum}_{1leq i <j<kleq n}ijkbigg)$ help me please

algebra-precalculus

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asked yesterday

jackyjacky

960615

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Using Newton's identities, you can express the sum at hand in terms of $sum_{k=1}^n k^ell$ for $ell = 1, 2, 3$...
  $endgroup$
  – achille hui
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @achille hui did not understand last line.please explain me thanks
  $endgroup$
  – jacky
  yesterday
  

  
  
  
  

add a comment | 

0

0

$begingroup$

Finding Coefficient of $x^{n-3}$ in $displaystyle prod^{n}_{k=1}(x-k)$

what i try

$displaystyle prod^{n}_{k=1}(x-k)=x^n-bigg(sum^{n}_{i=1}ibigg)+bigg(mathop{sumsum}_{1leq i <j leq n}ijbigg)x^{n-2}-bigg(mathop{sumsumsum}_{1leq i <j<kleq n}ijkbigg)x^{n-3}+cdots $

How do i find $displaystyle bigg(mathop{sumsumsum}_{1leq i <j<kleq n}ijkbigg)$ help me please

algebra-precalculus

share|cite|improve this question

asked yesterday

jackyjacky

960615

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Using Newton's identities, you can express the sum at hand in terms of $sum_{k=1}^n k^ell$ for $ell = 1, 2, 3$...
  $endgroup$
  – achille hui
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @achille hui did not understand last line.please explain me thanks
  $endgroup$
  – jacky
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Finding Coefficient of $x^{n-3}$ in $displaystyle prod^{n}_{k=1}(x-k)$

what i try

$displaystyle prod^{n}_{k=1}(x-k)=x^n-bigg(sum^{n}_{i=1}ibigg)+bigg(mathop{sumsum}_{1leq i <j leq n}ijbigg)x^{n-2}-bigg(mathop{sumsumsum}_{1leq i <j<kleq n}ijkbigg)x^{n-3}+cdots $

How do i find $displaystyle bigg(mathop{sumsumsum}_{1leq i <j<kleq n}ijkbigg)$ help me please

algebra-precalculus

share|cite|improve this question

asked yesterday

jackyjacky

960615

$endgroup$

share|cite|improve this question

asked yesterday

jackyjacky

960615

share|cite|improve this question

asked yesterday

jackyjacky

960615

asked yesterday

jackyjacky

960615

asked yesterday

jackyjacky

960615

1 Answer
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votes

2

$begingroup$

For every positive integer $n$, let $c_{n,j}=0$ if $j<0$ or $j>n$ and
$$prod_{k=1}^n(x+k)=sum_{j=0}^nc_{n,j}x^{n-j}$$
Then
begin{align}
sum_{j=0}^{n+1}c_{n+1,j}x^{n+1-j}
&=prod_{k=1}^{n+1}(x+k)\
&=(x+n+1)prod_{k=1}^{n}(x+k)\
&=(x+n+1)sum_{j=0}^nc_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=0}^n(n+1)c_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=1}^{n+1}(n+1)c_{n,j-1}x^{n+1-j}\
&=sum_{j=0}^n(c_{n,j}+(n+1)c_{n,j-1})x^{n+1-j}
end{align}
from which
$$c_{n+1,j}=c_{n,j}+(n+1)c_{n,j-1}$$
hence
begin{align}
c_{N,j}-c_{1,j}
&=sum_{n=1}^{N-1}(c_{n+1,j}-c_{n,j})\
&=sum_{n=1}^{N-1}(n+1)c_{n,j-1}
end{align}
that's
$$c_{N,j}=c_{1,j}+sum_{n=1}^{N-1}(n+1)c_{n,j-1}$$
Since $c_{n,0}=1$, we get
begin{align}
c_{N,1}
&=c_{1,1}+sum_{n=1}^{N-1}(n+1)c_{n,0}\
&=1+sum_{n=1}^{N-1}(n+1)\
&=sum_{n=1}^N n\
&=frac 12 N(N+1)\
c_{N,2}
&=c_{1,2}+sum_{n=1}^{N-1}(n+1)c_{n,1}\
&=frac 12sum_{n=1}^{N-1}n(n+1)^2\
&=frac 1{24}(N - 1) N (N + 1) (3 N + 2)\
c_{N,3}
&=c_{1,3}+sum_{n=1}^{N-1}(n+1)c_{n,2}\
&=frac 1{24}sum_{n=1}^{N-1}(n - 1) n (n + 1)^2 (3 n + 2)\
&=frac 1{48}(N - 2) (N - 1) N^2 (N + 1)^2
end{align}

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answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426

$endgroup$

add a comment | 

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2

$begingroup$

For every positive integer $n$, let $c_{n,j}=0$ if $j<0$ or $j>n$ and
$$prod_{k=1}^n(x+k)=sum_{j=0}^nc_{n,j}x^{n-j}$$
Then
begin{align}
sum_{j=0}^{n+1}c_{n+1,j}x^{n+1-j}
&=prod_{k=1}^{n+1}(x+k)\
&=(x+n+1)prod_{k=1}^{n}(x+k)\
&=(x+n+1)sum_{j=0}^nc_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=0}^n(n+1)c_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=1}^{n+1}(n+1)c_{n,j-1}x^{n+1-j}\
&=sum_{j=0}^n(c_{n,j}+(n+1)c_{n,j-1})x^{n+1-j}
end{align}
from which
$$c_{n+1,j}=c_{n,j}+(n+1)c_{n,j-1}$$
hence
begin{align}
c_{N,j}-c_{1,j}
&=sum_{n=1}^{N-1}(c_{n+1,j}-c_{n,j})\
&=sum_{n=1}^{N-1}(n+1)c_{n,j-1}
end{align}
that's
$$c_{N,j}=c_{1,j}+sum_{n=1}^{N-1}(n+1)c_{n,j-1}$$
Since $c_{n,0}=1$, we get
begin{align}
c_{N,1}
&=c_{1,1}+sum_{n=1}^{N-1}(n+1)c_{n,0}\
&=1+sum_{n=1}^{N-1}(n+1)\
&=sum_{n=1}^N n\
&=frac 12 N(N+1)\
c_{N,2}
&=c_{1,2}+sum_{n=1}^{N-1}(n+1)c_{n,1}\
&=frac 12sum_{n=1}^{N-1}n(n+1)^2\
&=frac 1{24}(N - 1) N (N + 1) (3 N + 2)\
c_{N,3}
&=c_{1,3}+sum_{n=1}^{N-1}(n+1)c_{n,2}\
&=frac 1{24}sum_{n=1}^{N-1}(n - 1) n (n + 1)^2 (3 n + 2)\
&=frac 1{48}(N - 2) (N - 1) N^2 (N + 1)^2
end{align}

share|cite|improve this answer

answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426

$endgroup$

add a comment | 

2

$begingroup$

For every positive integer $n$, let $c_{n,j}=0$ if $j<0$ or $j>n$ and
$$prod_{k=1}^n(x+k)=sum_{j=0}^nc_{n,j}x^{n-j}$$
Then
begin{align}
sum_{j=0}^{n+1}c_{n+1,j}x^{n+1-j}
&=prod_{k=1}^{n+1}(x+k)\
&=(x+n+1)prod_{k=1}^{n}(x+k)\
&=(x+n+1)sum_{j=0}^nc_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=0}^n(n+1)c_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=1}^{n+1}(n+1)c_{n,j-1}x^{n+1-j}\
&=sum_{j=0}^n(c_{n,j}+(n+1)c_{n,j-1})x^{n+1-j}
end{align}
from which
$$c_{n+1,j}=c_{n,j}+(n+1)c_{n,j-1}$$
hence
begin{align}
c_{N,j}-c_{1,j}
&=sum_{n=1}^{N-1}(c_{n+1,j}-c_{n,j})\
&=sum_{n=1}^{N-1}(n+1)c_{n,j-1}
end{align}
that's
$$c_{N,j}=c_{1,j}+sum_{n=1}^{N-1}(n+1)c_{n,j-1}$$
Since $c_{n,0}=1$, we get
begin{align}
c_{N,1}
&=c_{1,1}+sum_{n=1}^{N-1}(n+1)c_{n,0}\
&=1+sum_{n=1}^{N-1}(n+1)\
&=sum_{n=1}^N n\
&=frac 12 N(N+1)\
c_{N,2}
&=c_{1,2}+sum_{n=1}^{N-1}(n+1)c_{n,1}\
&=frac 12sum_{n=1}^{N-1}n(n+1)^2\
&=frac 1{24}(N - 1) N (N + 1) (3 N + 2)\
c_{N,3}
&=c_{1,3}+sum_{n=1}^{N-1}(n+1)c_{n,2}\
&=frac 1{24}sum_{n=1}^{N-1}(n - 1) n (n + 1)^2 (3 n + 2)\
&=frac 1{48}(N - 2) (N - 1) N^2 (N + 1)^2
end{align}

share|cite|improve this answer

answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426

$endgroup$

add a comment | 

$begingroup$

For every positive integer $n$, let $c_{n,j}=0$ if $j<0$ or $j>n$ and
$$prod_{k=1}^n(x+k)=sum_{j=0}^nc_{n,j}x^{n-j}$$
Then
begin{align}
sum_{j=0}^{n+1}c_{n+1,j}x^{n+1-j}
&=prod_{k=1}^{n+1}(x+k)\
&=(x+n+1)prod_{k=1}^{n}(x+k)\
&=(x+n+1)sum_{j=0}^nc_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=0}^n(n+1)c_{n,j}x^{n-j}\
&=sum_{j=0}^nc_{n,j}x^{n+1-j}+sum_{j=1}^{n+1}(n+1)c_{n,j-1}x^{n+1-j}\
&=sum_{j=0}^n(c_{n,j}+(n+1)c_{n,j-1})x^{n+1-j}
end{align}
from which
$$c_{n+1,j}=c_{n,j}+(n+1)c_{n,j-1}$$
hence
begin{align}
c_{N,j}-c_{1,j}
&=sum_{n=1}^{N-1}(c_{n+1,j}-c_{n,j})\
&=sum_{n=1}^{N-1}(n+1)c_{n,j-1}
end{align}
that's
$$c_{N,j}=c_{1,j}+sum_{n=1}^{N-1}(n+1)c_{n,j-1}$$
Since $c_{n,0}=1$, we get
begin{align}
c_{N,1}
&=c_{1,1}+sum_{n=1}^{N-1}(n+1)c_{n,0}\
&=1+sum_{n=1}^{N-1}(n+1)\
&=sum_{n=1}^N n\
&=frac 12 N(N+1)\
c_{N,2}
&=c_{1,2}+sum_{n=1}^{N-1}(n+1)c_{n,1}\
&=frac 12sum_{n=1}^{N-1}n(n+1)^2\
&=frac 1{24}(N - 1) N (N + 1) (3 N + 2)\
c_{N,3}
&=c_{1,3}+sum_{n=1}^{N-1}(n+1)c_{n,2}\
&=frac 1{24}sum_{n=1}^{N-1}(n - 1) n (n + 1)^2 (3 n + 2)\
&=frac 1{48}(N - 2) (N - 1) N^2 (N + 1)^2
end{align}

share|cite|improve this answer

answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426

$endgroup$

share|cite|improve this answer

answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426

answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426

answered 19 hours ago

Fabio LucchiniFabio Lucchini

8,62811426