2016 is coming!

Nice! We have not seen one of those in a while...

Let $n=2016$ and substitute $y=\frac{1}{1-x}$ or $x=\frac{y-1}{y}$ . Since the $x_j$ satisfy the equation $x^{n+1}=1$ , the corresponding $y_j$ will satisfy $(y-1)^{n+1}=y^{n+1}$ , or, simplified, $y^n-\frac{n}{2}y^{n-1}+\frac{n(n-1)}{6}y^{n-2}-...=0$ . It follows from Viete that $\sum_{j=1}^{n}y_j^2=\left(\frac{n}{2}\right)^2-\frac{2n(n-1)}{6}=\frac{n(4-n)}{12}$ . For $n=2016$ this comes out to be $\boxed{-338016}$

This method is less elegant, but involves one of my favourite formulae! If we write $\zeta = e^{\frac{2\pi i}{2n+1}}$ , then $\begin{array}{rcl} S \; =\; \displaystyle\sum_{j=1}^{2n} \dfrac{1}{(1 - \zeta^j)^2} & = & \displaystyle\sum_{j=1}^{2n} \dfrac{1}{\zeta^{2j} + 1 - 2\zeta^j} \; = \; \sum_{j=1}^{2n} \frac{\zeta^{-j}}{2\cos \frac{2\pi j}{2n+1} - 2} \; = \; \displaystyle-\frac14\sum_{j=1}^{2n} \frac{\zeta^{-j}}{\sin^2\frac{\pi j}{2n+1}} \end{array}$ Replacing $j$ by $2n+1-j$ , this last sum is also equal to $-\dfrac{1}{4}\sum_{j=1}^{2n} \frac{\zeta^{j}}{\sin^2\frac{\pi j}{2n+1}}$ and hence $\begin{array}{rcl} S & = & \displaystyle -\dfrac{1}{4}\sum_{j=1}^{2n}\frac{\cos\frac{2\pi j}{2n+1}}{\sin^2\frac{\pi j}{2n+1}} \; = \; -\dfrac{1}{4}\sum_{j=1}^{2n}\frac{1 - 2\sin^2\frac{\pi j}{2n+1}}{\sin^2\frac{\pi j}{2n+1}} \; = \; n - \displaystyle\dfrac14\sum_{j=1}^{2n} \mathrm{cosec}^2\,\frac{\pi j}{2n+1} \\ & = & \displaystyle \tfrac12n - \dfrac12\sum_{j=1}^n \cot^2\frac{\pi j}{2n+1} \end{array}$ Now $\begin{array}{rcl} \sin(2n+1)\theta & = & \displaystyle\mathrm{Im}\big[(\cos\theta + i\sin\theta)^{2n+1}\big] \; = \; \sum_{j=0}^n {2n+1 \choose 2j+1}(-1)^j \cos^{2n-2j}\theta \sin^{2j+1}\theta \\ & = & \displaystyle\sin^{2n+1}\theta \sum_{j=0}^n {2n+1 \choose 2j+1}(-1)^j \cot^{2n-2j}\theta \; = \; \sin^{2n+1}\theta f_n(\cot^2\theta) \end{array}$ where $f_n(X) \; = \; \sum_{j=0}^n {2n+1 \choose 2j+1} (-1)^j X^{n-j} \;.$ The roots of $f_n(X)$ are thus $\cot^2\dfrac{\pi j}{2n+1}$ for $1 \le j \le n$ and hence $\sum_{j=1}^n \cot^2\dfrac{\pi j}{2n+1} \; = \; {2n+1 \choose 3} {2n+1 \choose 1}^{-1} \; = \; \dfrac13n(2n-1)$ This is the formula I always think remarkable! Thus $S \; = \; \tfrac12n - \tfrac16n(2n-1) \; = \; -\tfrac13n(n-2) \;.$ For $n = 1008$ , the answer is $S = -338106$ .

$P(x)=\left(x-x_{1}\right)\left (x-x_{2} \right)....\left (x-x_{2016} \right)$

$\log P(x)=\log \left(x-x_{1}\right) +\log \left (x-x_{2} \right) +.... +\log \left (x-x_{2016} \right)$

Taking derivative of both sides

$\frac{1}{P(x)} \frac{dy}{dx} = \frac{1}{1-x_{1}} + \frac{1}{1-x_{2}}+...+\frac{1}{1-x_{2016}}$

Differentiate once again

$\frac{P^{''} (x)}{P(x)}-\left(\frac{P^{'} (x)}{P(x)}\right)^2 = -\left[\frac{1}{(1-x_{1})^2} + \frac{1}{(1-x_{2})^2}+...+\frac{1}{(1-x_{2016})^2}\right]$

Multiply by $-1$

$\left(\frac{P^{'} (x)}{P(x)}\right)^2-\frac{P^{''} (x)}{P(x)} = \left[\frac{1}{(1-x_{1})^2} + \frac{1}{(1-x_{2})^2}+...+\frac{1}{(1-x_{2016})^2}\right]$

Put $x=1$ in above equation and thats the desired answer

$P(1)=2017, P'(x)=2016x^{2015}+2015x^{2014}+...+2x+1$ so $P'(1)=2016+2015+...1=\frac{(2016)(2015)}{2}$

$P''(x)=(2016)(2015)x^{2014}+(2015)(2014)x^{2013}+...+(3)(2)x+(2)(1)$

$P''(1)=\displaystyle \sum_{n=1}^{2015} n(n+1) =\displaystyle \sum_{n=1}^{2015} n^2+ \displaystyle \sum_{n=1}^{2015} n = \frac{(2015)(2016)(4031)}{6}+ \frac{(2015)(2016)}{2} = \frac{(2015)(2016)}{6}\left(4031+3\right) = \frac{(2015)(2016)}{6}\left(4034\right)$

$\frac{P'(1)}{P(1)}=\frac{(2016)(2017)}{2 \times 2017} = \frac{2016}{2}= 1008$

$\frac{P''(1)}{P(1)}=\frac{(2015)(2016)}{6}\left(\frac{4034}{2017}\right)=\frac{(2015)(2016)}{6} \times 2 = 2015 \times 672$

Now Required answer = $\left(\frac{P'(1)}{P(1)}\right)^2 - \frac{P''(1)}{P(1)} = (1008)^2 - 2015\times 6072 = \boxed{-338016}$

Given equation :
$\displaystyle \sum_{r=0}^{2016}x^{r} = 0$
Let $y = \dfrac{1}{1-x} \rightarrow x = \dfrac{y-1}{y}$
Substituting in the given equation,
$\displaystyle \sum_{r=0}^{2016}\left(\dfrac{y-1}{y}\right)^{r} = 0$
Multiplying by $y^{2016}$
$\displaystyle \sum_{r=0}^{2016}(y-1)^{r}y^{2016-r} = 0$
$y^{2017} - (y-1)^{2017} = \left(y-(y-1)\right) \times \left( \displaystyle \sum_{r=0}^{r=2016}(y-1)^{r}y^{2016-r}\right)$
Using the above relation,

$(y)^{2017} - (y-1)^{2017} = 0$
Co-efficient of $y^{2016}$ (a) = $\dbinom{2017}{1}$
Co-efficient of $y^{2015}$ (b)= $\displaystyle - \dbinom{2017}{2}$
Co-efficient of $y^{2014}$ (c) = $\dbinom{2017}{3}$
Let $S_{i}$ denote the symmetric sums of roots of the polynomial.
According to Vieta's,
$S_{1} = \dfrac{-b}{a} = \dfrac{\dbinom{2017}{2}}{\dbinom{2017}{1}} = 1008$
$S_{2} = \dfrac{c}{a} = \dfrac{\dbinom{2017}{3}}{\dbinom{2017}{1}} =\dfrac{2016\times2015}{6}$
Applying Newton's identity,
$P_{2} = S_{1}^2 - 2S_{2}$
$P_{2} = 1008^2 - \dfrac{2016\times2015}{3}$
$P_{2} = 1016064 - 1354080 = -338016$

I used a calculus approach by labeling the given polynomial as $p(x).$ Taking derivatives twice gives

$p''(x) = p'(x)t(x) - p(x) s(x)$

where $t$ is the sum of the reciprocals of the linear factors of $p$ and $s$ is the sum of squared reciprocals of linear factors of $p$ . Evaluate everything at $x=1$ and $s(1)$ gives the desired value. :)

Using an analytical way with the help of Vieta's formula I got a closed formula for n (highest coefficient):

$\frac{1}{n+1}[\frac{n^{2}(n+1)}{4}-2\sum_{i=0}^{n-2}\frac{(n-i)!}{2!(n-i-2)!}]$

Simplified this expression results in:

$\frac{n}{12}(4-n)$