The Degree Isn't Given Either?

$\displaystyle \sum_{ k = 1}^n kx^k = n\cdot\prod_{k = 1}^n (x - a_k) \Rightarrow \ln\left(\sum_{ k = 1}^n kx^k \right) = \ln n + \sum_{k =1}^n \ln(x - a_k) = \ln \left( n \cdot \prod_{k = 1}^n (x - a_k) \right) \Rightarrow$ Derivating with respect to $x$ $\sum_{k =1}^n \frac{1}{x - a_k} = \frac{\displaystyle \sum_{k =1}^n k^2x^{k -1}}{\displaystyle \sum_{ k = 1}^n kx^k}$ Substituting $x = 1$ $\sum_{k =1}^n \frac{1}{1 - a_k} = \frac{\displaystyle \sum_{k =1}^n k^2\cdot1^{k -1}}{\displaystyle \sum_{ k = 1}^n k\cdot1^k} = 7 = \frac{\frac{n(n +1)(2n +1)}{6}}{\frac{n(n +1)}{2}} \Rightarrow$ $\frac{2n + 1}{3} = 7 \Rightarrow n = 10$

Note .- If $\displaystyle p(x) = \sum_{ k = 1}^n kx^k$ then $x = 1$ is not a root of $p(x)$ because $\sum_{ k = 1}^n k\cdot1^k = \frac{n(n + 1)}{2} \neq 0$

Since $a_i$ are the roots to the equation $\sum k x^k = 0$ , using the transformation $y = \frac{1}{1-x}$ or $x = 1 - \frac{1}{y} = \frac{y-1}{y}$ , we get that $\frac{1}{1- a_i}$ are the roots to

$\sum k x^k = 0 \Leftrightarrow \sum k \left( \frac{ y-1} {y} \right)^k = 0 \Leftrightarrow \sum k y^{n-k} (y-1)^k = 0$

Now, we want to find $\sum \frac{1}{1-a_i}$ , which are the sum of the roots to the equation in $y$ . Hence, we need to find the negative coefficient of $y^{n-1}$ divided by the coefficient of $y^{n}$ , which is just:

$\frac{ - \sum k \times { k \choose 1 } \times (-1)^1 } { \sum k \times 1 } = \frac{ \sum k^2 }{ \sum k } = \frac{ \frac{n(n+1)(2n+1) } {6} } { \frac{n(n+1) } { 2} } = \frac{2n+1}{3}$

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

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. $f(1) = \lim_{x\to 1}\dfrac{x\left(nx^{n+1}-(n+1)x^n + 1\right)}{(x-1)^2} = \sum_0^n i = \dfrac{n(n+1)}2$

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

Using formula, $\displaystyle \sum_1^n \dfrac1{1-a_i} = \dfrac{f'(1)}{f(1)}$ , we get $7=\dfrac{2n+1}3$

So, $\boxed{n = 10}$

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Proof for the above formula,

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

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

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

The polynomial $\frac{\sum_{k=1}^n k(1+x)^k}{\sum_{k=1}^n k}$ has $a_k-1$ as its roots. Since the constant term is $1$ , the polynomial can be factored as $\Big(1-\frac{x}{a_1-1}\Big)\Big(1-\frac{x}{a_2-1}\Big)...(1-\frac{x}{a_n-1}\Big)$ . If one expands this product, then she finds that the coefficient of the linear term is equal to $\sum_{k=1}^n \frac{1}{1-a_k}$ . (One can also note that the coefficient of the linear term is the same as the coefficient of $x^{n-1}$ .) This means that the coefficient of the linear term must equal $7$ . The coefficient of the linear term also has a representation in terms of $n$ . If one applies the binomial theorem to $(1+x)^k$ , she can determine the coefficient of the linear term as $\frac{\sum k^2}{\sum k}$ . (I'll leave that to the reader.) Hence, $\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=7\Rightarrow \frac{2n+1}{3}=7\Rightarrow n=10$ .