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The title says most of it. I have found that the sum of the coefficients of the polynomial$(x-1)(x-2)(x-3)\cdots(x-(n-1))$ yields $n!$.

For example, the coefficients of the polynomial $(x-1)(x-2)$ sum to $1+3+2=6=3!$ and similarly the coefficients of $(x-1)(x-2)(x-3)$ sum to $1+6+11+6=24=4!$.

My question, then is how might I prove this more generally. I have the feeling that induction might be an optimal way to go about this, but am unsure of the specifics. Any help is appreciated!

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  • $\begingroup$ Induction on degree. $\endgroup$ Feb 8, 2020 at 2:43
  • $\begingroup$ product of arithmetic progression in terms of the middle term ... $\endgroup$
    – user645636
    Feb 8, 2020 at 2:46

3 Answers 3

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Hint: Let

$$f(x) = (x-1)(x-2)(x-3)...(x-(n-1)) \tag{1}\label{eq1A}$$

Note $f(-1) = (-2)(-3)\cdots(-n) = (-1)^{n-1}(n!)$. Consider what this value is in the expansion of $f(x)$ as a polynomial in $x$, plus how it relates to the sum of the magnitudes of the coefficients.

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  • $\begingroup$ Thank you! Should have thought to consider f(-1)... Seems somewhat like how you show the nth row of Pascal’s triangle sums to 2^n $\endgroup$
    – nak17
    Feb 8, 2020 at 2:49
  • $\begingroup$ @nak17 You're welcome. Yes, it's roughly similar to one way to get that the sum of the $n$'th row of Pascal's triangle is $2^n$. $\endgroup$ Feb 8, 2020 at 2:52
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The only reason why you're adding up the magnitudes of the coefficients, and not the coefficients themselves, is that some of them are negative. If instead of considering $(x-1)(x-2)\dots(x-n)$ we consider $(x+1)(x+2)\dots(x+n)$, we'll be left with a polynomial whose coefficients have the same magnitudes, but are always positive. So let's do that.

Now, for any polynomial $P(x)$, $P(1)$ gives the sum of the coefficients: if $P(x)=a_0+a_1x+a_2x^2+\dots+a_nx^n$, then $P(1)=a_0+a_1(1)+a_2(1^2)+\dots+a_n(1^n)=a_0+a_1+a_2+\dots+a_n$.

So let $P_n(x)=(x+1)(x+2)\dots(x+n)$. Then

\begin{align*} P_n(1)&=(1+1)(1+2)\dots(1+n)\\ &=2(3)\dots(n+1)\\ &=(n+1)! \end{align*}

That is, the sum of the coefficients of $P_n$ is $(n+1)!$, as you've observed.

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Since all roots of the polynomial are positive, Descartes' Rule of signs guarantees that the signs of the coefficients alternate, for example with $n=4$:

$(x-1)(x-2)(x-3)=x^3\color{purple}{-}6x^2\color{purple}{+}11x\color{purple}{-}6$

Because of this sign pattern the absolute values of the coefficients are recovered by putting $x=-1$ so that the powers of $x$ correlate with the alternating signs of the coefficients:

$(-1-1)(-1-2)(-1-3)=(-1)^3-6(-1)^2+11(-1)11x-6=-(1+6+11+6)$

and upon taking absolute values

$(1+1)(1+2)(1+3)=1+6+11+6$

where for the general case the product on the left is $n!$ and the sum on the right contains the desired coefficient magnitudes.

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