# Sum of magnitudes of coefficients of polynomial $(x-1)(x-2)(x-3)\cdots(x-(n-1))$

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!

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

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.

• 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 Feb 8, 2020 at 2:49
• @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 $2^n$. Feb 8, 2020 at 2:52

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.

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.