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There is a question I am having difficult time with. Please help me to find the coefficient of $x^8$ in the polynomial $$(x-1)(x-2)(x-3)........(x-10)$$ I don't even know to start but I think it's related to binomial theorem but again there are no powers here. Any help is appreciated. Thank you.

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  • $\begingroup$ here is your term that i have expanded $${x}^{10}-55\,{x}^{9}+1320\,{x}^{8}-18150\,{x}^{7}+157773\,{x}^{6}- 902055\,{x}^{5}+3416930\,{x}^{4}-8409500\,{x}^{3}+12753576\,{x}^{2}- 10628640\,x+3628800 $$ $\endgroup$ – Dr. Sonnhard Graubner Dec 11 '16 at 14:51
  • $\begingroup$ Yes, you got the right answer i.e 1320 but can you please explain how you got it. Help will be appreciated. $\endgroup$ – Mayank Mittal Dec 11 '16 at 14:57
  • $\begingroup$ @Dr.SonnhardGraubner WolframAlpha? $\endgroup$ – Simply Beautiful Art Dec 11 '16 at 14:59
  • $\begingroup$ If you know Taylor's theorem, just take the $8$th derivative $\endgroup$ – Simply Beautiful Art Dec 11 '16 at 15:21
  • $\begingroup$ See Stirling numbers of the first kind. $\endgroup$ – Lucian Dec 19 '16 at 0:25
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The coefficient of $x^8$ of $\prod_{k=1}^{10}(x-k)$ is given by $$\sum_{1\leq j<k\leq 10}kj=\frac{1}{2}\left(\left(\sum_{k=1}^{10}k\right)^2-\sum_{k=1}^{10}k^2\right)= \frac{1}{2}\left(55^2-385\right)=1320.$$ See also Vieta's formulas.

In other words, here the coefficient of $x^8$ is given by the sum of all products of two distinct numbers in $\{1,2,..,10\}$. This sum can be obtained by squaring $(1+2+3+\dots+10)$ and then by throwing away all the squares $1,4,\dots, 100$. The result should be divided by 2 (they are double products).

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  • $\begingroup$ Vieta's formulas? $\endgroup$ – Simply Beautiful Art Dec 11 '16 at 14:59
  • $\begingroup$ @Simple Art Yes, that is a good reference. $\endgroup$ – Robert Z Dec 11 '16 at 15:00
  • $\begingroup$ Thank you for your help but I am in just class 9 and we have not been taught summation yet so can you please post a simpler way out. $\endgroup$ – Mayank Mittal Dec 11 '16 at 15:04
  • $\begingroup$ It is just a notation. $\sum_{k=1}^{10}k=(1+2+3+\dots+10)$ and $\sum_{k=1}^{10}k^2=(1+4+9+\dots+100)$ $\endgroup$ – Robert Z Dec 11 '16 at 15:06
  • $\begingroup$ @infinitely curious See my edited answer. Is it clear now? $\endgroup$ – Robert Z Dec 11 '16 at 15:16
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First, we see how we get the powers of $x^8$ thru the traditional method of "expanding out".

To get powers of $x^8$, we will actually multiply 8 $x$'s together with two numbers, e.g.

$x\dots x\cdot (-2)(-3)=6x^8$

So this boils down to finding all the sums of products of two distinct numbers, i.e. $1\times 2+1\times3+\dots+9\times 10$.

First we find out what is $1(2)+1(3)+1(4)+\dots+1(10)$. Surely, it is 54. One other way to write it is $(1+2+\dots+10)-1=54$

Next, we find the product of two distinct numbers, containing 2, i.e.: $2(2)+2(3)+2(4)+\dots+2(10)$. This is $2(1+3+\dots+10)=2(1+2+\dots+10)-2^2$.

The shortcut to this is to observe that the sum is actually:

$(1+2+\dots+10)(1+2+\dots+10)-(1^2+2^2+\dots+10^2)=55(55)-385=2640$

Note that we have double counted: for instance $1\times 2$ and $2\times 1$ are counted twice.

So dividing by 2 gives 1320 as desired.

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    $\begingroup$ i understood both the answers( both are more or less the same thing ) but this is for $x^8$. How will we find coefficient of $x^9$ or something else. Will there be a distinct method ? $\endgroup$ – Mayank Mittal Dec 11 '16 at 15:33
  • $\begingroup$ @infinitelycurious The method will be similar. In this case it is even easier, it is the sum of the distinct (negative) numbers, i.e. $-(1+2+\dots+10)=-55$. $\endgroup$ – yoyostein Dec 11 '16 at 15:42
  • $\begingroup$ But on what basis do you determine what is to be taken to find the coefficient. Like for $x^8$ you took sum of the products of all the distinct numbers and for $x^9$ you took sum of the distinct numbers. Thank you $\endgroup$ – Mayank Mittal Dec 11 '16 at 15:55
  • $\begingroup$ @infinitelycurious For $x^9$, we multiply 9 $x$'s together with 1 number. For $x^8$, we multiply 8 $x$'s together with 2 numbers. You can try expanding the expression a bit to get the idea. $\endgroup$ – yoyostein Dec 11 '16 at 16:10

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