Inspired by Mr. $\pi$ (The force awakens)

If $f(X)=\sum_{k=1}^n kX^k$ has roots $a_j$ for $1\le j\le n$ , then putting $X=1-Y^{-1}=\tfrac{Y-1}{Y}$ yields $g(Y)=\sum_{k=1}^n k(Y-1)^kY^{n-k}$ as the polynomial with roots $(1-a_j)^{-1}$ for $1\le j\le n$ . Now $\begin{array}{rcl} g(Y) &=& \displaystyle Y^n \sum_{k=1}^n k -Y^{n-1} \sum_{k=1}^n k^2 + Y^{n-2} \sum_{k=1}^n k\binom{k}{2} + ... \\ &=& \displaystyle \tfrac12n(n+1)Y^n - \tfrac16n(n+1)(2n+1)Y^{n-1} + \tfrac1{24}n(n+1)(n-1)(3n+2)Y^{n-2}+... \end{array}$ so $\sum_j \frac{1}{1-a_j} = \tfrac13(2n+1) \qquad \sum_{j<k}\frac{1}{(1-a_j)(1-a_k)}=\tfrac{1}{12}(n-1)(3n+2)$ and so $\sum_j \frac{1}{(1-a_j)^2}=\tfrac19(2n+1)^2 -\tfrac16(n-1)(3n+2)=\tfrac{1}{18}(8+11n-n^2)$ We need to solve $\tfrac{1}{18}(8+11n-n^2)=-13$ , which factorises as $(n-22)(n+11)=0$ , and so $n=\boxed{22}$ .

Let $f(x) = \sum_1^n ix^i =n\prod_1^n (x-x_i)$

We know, $\sum_0^n x^i = \dfrac{x^{n+1}-1}{x-1}\\\dfrac{d}{dx}\sum_0^n x^i = \sum_0^n ix^{i-1}$

So, $f$ becomes, $\begin{aligned}f(x) &= x \dfrac{d}{dx}\dfrac{x^{n+1}-1}{x-1}\\&= \dfrac{x\left(nx^{n+1}-(n+1)x^n + 1\right)}{(x-1)^2}\end{aligned}$

We know $1$ is not a root. So, we can use limit value to get $f(1)$ or use sum of A.P formula. $f(1) = \lim_{x\to 1}\dfrac{x\left(nx^{n+1}-(n+1)x^n + 1\right)}{(x-1)^2} = \sum_0^n i = \dfrac{n(n+1)}2$

$f'(1) = \sum_0^n i^2 = \dfrac{n(n+1)(2n+1)}6$

$f''(1) = \lim_{x\to1} \dfrac{d}{dx}\left(\dfrac{d}{dx}\left[\dfrac{x\left(nx^{n+1}-(n+1)x^n + 1\right)}{(x-1)^2}\right]\right) = \dfrac{(n-1)n(n+1)(3n+2)}{24}\\\text{OR}\\f''(1) = \sum_1^n k \binom{k}{2} = \dfrac{(n-1)n(n+1)(3n+2)}{24}$

Using formula, $\displaystyle\sum_1^n \dfrac1{(1-x_i)^2} = \dfrac{(f'(1))^2 - f''(1)f(1)}{(f(1))^2}$ , we get $-13 = \dfrac{-n^2+11n+8}{18}$

So, $n= -11,22$

Taking positive value, $\boxed{n=22}$

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Proof for the above formula,

Suppose $f(x) = k\prod_1^n (x-x_i)$

Differentiating once, $f'(x) = \sum_1^n \dfrac{f(x)}{x-x_i} =f(x)\sum_1^n \dfrac1{x-x_i}$

So, $\sum_1^n \dfrac1{x-x_i} = \dfrac{f'(x)}{f(x)}$

Differentiating again, $\begin{aligned}f''(x) &= \sum_1^n \left[\dfrac{f'(x)}{x-x_i} - \dfrac{f(x)}{(x-x_i)^2}\right]\\ &= f'(x)\cdot \dfrac{f'(x)}{f(x)} - f(x)\sum_1^n \dfrac1{(x-x_i)^2}\end{aligned}\\ \Rightarrow \boxed{\displaystyle \sum_1^n \dfrac1{(x-x_i)^2} = \dfrac{(f'(x))^2 - f''(x)f(x)}{(f(x))^2}}$

I came up with a technique to solve this problem, but after trying higher and higher degree polynomials, I eventually gave up and used Matlab. So I figured I better redeem myself by posting a real solution based on the technique I came up with. I will use my own method to derive a polynomial in $n$ and find the roots, as in the other solutions. I divided by $x$ to get rid of the zero root (call it $a_1$ ). Therefore, I was looking for when $\sum_{k=2}^n\frac{1}{(1-a_k)^2}=-14$ . I started by creating two polynomials of degree $n-1$ , one with roots $1-a_k$ ( $2\leq k\leq n$ ), and the other with roots $a_k-1$ ( $2\leq k \leq n$ ). Respectively, these polynomials take the form $P_1(x)=\sum_{k=1}^n k(1-x)^{k-1}$ and $P_2(x)=\sum_{k=1}^n k(1+x)^{k-1}$ . I can factorize them in the following way: $P_1(x)=\Big(\sum_{k=1}^n k\Big)\Big(1-\frac{x}{1-a_2}\Big)\Big(1-\frac{x}{1-a_3}\Big)...\Big(1-\frac{x}{1-a_n}\Big)$ and $P_2(x)=\Big(\sum_{k=1}^n k \Big)\Big(1+\frac{x}{1-a_2}\Big)\Big(1+\frac{x}{1-a_3}\Big)...\Big(1+\frac{x}{1-a_n}\Big)$ The product of these two polynomials can be written as $P_1(x)P_2(x)=\Big(\sum_{k=1}^n k\Big)^2\Big(1-\frac{x^2}{(1-a_2)^2}\Big)\Big(1-\frac{x^2}{(1-a_3)^2}\Big)...\Big(1-\frac{x^2}{(1-a_n)^2}\Big)$ The $x^2$ coefficient is given by $\Big(\sum_{k=1}^n k\Big)^2 \Big(-\sum_{k=2}^n \frac{1}{(1-a_k)^2}\Big)=14\Big(\sum_{k=1}^n k\Big)^2$ . Now consider each polynomial in standard form (using the binomial theorem) and take their product: $P_1(x)P_2(x)=\Big(\sum_{k=1}^n k - x\sum_{k=1}^n k(k-1) +x^2\sum_{k=1}^n \frac{k(k-1)(k-2)}{2} +\mathcal{O}(x^3) \Big) \Big( \sum_{k=1}^n k + x\sum_{k=1}^n k(k-1) +x^2\sum_{k=1}^n \frac{k(k-1)(k-2)}{2} +\mathcal{O}(x^3)\Big)$ From this I find another expression that represents the $x^2$ coefficient. Equating the two: $\Big(\sum_{k=1}^n k(k-1)(k-2)\Big)\Big(\sum_{k=1}^n k\Big)-\Big(\sum_{k=1}^n k(k-1) \Big)^2=14\Big(\sum_{k=1}^n k\Big)^2$ $\frac{n(n+1)}{2}\Big(\frac{n^2(n+1)^2}{4}-\frac{n(n+1)(2n+1)}{2}+n(n+1) \Big)-\Big(\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2} \Big)^2 =14\frac{n^2(n+1)^2}{4}$ Next, divide by $\frac{n^2(n+1)^2}{4}$ to obtain $\frac{n(n+1)}{2}-(2n+1)+2-\Big(\frac{2n+1}{3}-1\Big)^2=14$ This reduces to $n^2-11n-242=0$ which, has roots $n=-11, 22$ . Wait. -11??!... lol