The answer is $n^2$

As $1,\alpha_1, \cdots ,\alpha_4$ are roots of the equation $x^5-1=0,$ $\begin{aligned} (x-1)\prod_{i=1}^{4} (x-\alpha_i)&=x^5-1\\ \prod_{i=1}^{4} (x-\alpha_i)&=\dfrac{x^5-1}{x-1} \end{aligned}$

Take the natural logarithm on both the sides and then differentiate wrt $x$ .

$\begin{aligned} \sum_{i=1}^{4} \log(x-\alpha_i)&=\log(x^5-1)-\log(x-1)\\ \sum_{i=1}^{4} \dfrac{1}{x-\alpha_i}&=\dfrac{5x^4}{x^5-1}-\dfrac{1}{x-1} \end{aligned}$ Substitute $x=2$ in the above obtained equality to see that $\sum_{i=1}^{4} \dfrac{1}{2-\alpha_i}=\dfrac{49}{31}$ $\therefore \sum_{i=1}^{4} \dfrac{31}{2-\alpha_i}=\boxed{49}$

if $a_1,a_2,.....a_n$ are the roots of $f(x)=0$ , then $\dfrac{f'(x)}{f(x)}=\sum_{i=1}^n \dfrac{1}{x-a_i}.$

Considering $f(x)=x^5-1$ here and using the above property, we have

$\dfrac{5x^4}{x^5-1}=\dfrac{1}{x-1}+\dfrac{1}{x-\alpha_1}+\dfrac{1}{x-\alpha_2}+\dfrac{1}{x-\alpha_3}+\dfrac{1}{x-\alpha_4}.$

Putting $x=2$ we get

$\dfrac{80}{31}=\dfrac{1}{1}+\dfrac{1}{2-\alpha_1}+\dfrac{1}{2-\alpha_2}+\dfrac{1}{2-\alpha_3}+\dfrac{1}{2-\alpha_4}$

$\implies \dfrac{31}{2-\alpha_1}+\dfrac{31}{2-\alpha_2}+\dfrac{31}{2-\alpha_3}+\dfrac{31}{2-\alpha_4}=\boxed{49}.$

Calculus method is best & easiest but i am posting a different method.

$1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }=\left( x-{ \alpha }_{ 1 } \right) \left( x-{ \alpha }_{ 2 } \right) \left( x-{ \alpha }_{ 3 } \right) \left( x-{ \alpha }_{ 4 } \right) \\ \\ Let\quad x=2-t\quad \& 2-\alpha =\beta \\ \\ P=1+\left( 2-t \right) +{ \left( 2-t \right) }^{ 2 }+{ \left( 2-t \right) }^{ 3 }+{ \left( 2-t \right) }^{ 4 }\\ \\ \& \left( 2-t-{ \alpha }_{ 1 } \right) \left( 2-t-{ \alpha }_{ 2 } \right) \left( 2-t-{ \alpha }_{ 3 } \right) \left( 2-t-{ \alpha }_{ 4 } \right) \\ \\ \quad \quad \quad =\left( t-{ \beta }_{ 1 } \right) \left( t-{ \beta }_{ 2 } \right) \left( t-{ \beta }_{ 3 } \right) \left( t-{ \beta }_{ 4 } \right) \\ \\ \quad \quad \therefore P=\left( t-{ \beta }_{ 1 } \right) \left( t-{ \beta }_{ 2 } \right) \left( t-{ \beta }_{ 3 } \right) \left( t-{ \beta }_{ 4 } \right) \\ \\ \therefore \cfrac { 1 }{ { \beta }_{ 1 } } +\cfrac { 1 }{ { \beta }_{ 2 } } +\cfrac { 1 }{ { \beta }_{ 3 } } +\cfrac { 1 }{ { \beta }_{ 4 } } =\cfrac { \sum { { \beta }_{ 1 } } { \beta }_{ 2 }{ \beta }_{ 3 } }{ { \beta }_{ 1 }{ \beta }_{ 2 }{ \beta }_{ 3 }{ \beta }_{ 4 } } \\ \\ \quad \quad =\cfrac { -coff.\quad of\quad t\quad in\quad P }{ const.\quad term\quad in\quad P } \\ \\ \quad =\left( -1 \right) \cfrac { \left\{ -1-{ 2 }^{ 1 }\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) { -2 }^{ 2 }\left( \begin{matrix} 3 \\ 1 \end{matrix} \right) -{ 2 }^{ 3 }\left( \begin{matrix} 4 \\ 1 \end{matrix} \right) \right\} }{ 1+2+{ 2 }^{ 2 }+{ 2 }^{ 3 }+{ 2 }^{ 4 } } \\ \\ \quad =\cfrac { 49 }{ 31 } \\ \\ \therefore \cfrac { 31 }{ 2-{ \alpha }_{ 1 } } +\cfrac { 31 }{ 2-{ \alpha }_{ 2 } } +\cfrac { 31 }{ 2-{ \alpha }_{ 3 } } +\cfrac { 31 }{ 2-{ \alpha }_{ 4 } } =49$

The required expression can be rewritten as $\displaystyle\sum _{i = 1} ^{4} \dfrac{2^5 - \alpha_i ^5 }{2 - \alpha_i}$ .

We see that $\dfrac{x^5 - a^5}{x - a} = x^4 + ax^3 + a^2 x^2 + a^3 x + a^4$ . After substituting $x = 2$ and $a = \alpha_i$ , the required expression becomes $\displaystyle\sum _{i = 1} ^{4} (16 + 8 \alpha_i + 4 \alpha_i ^2 + 2 \alpha_i^3 + \alpha_i^4)$

Since $\displaystyle\sum _{i=1} ^{4} \alpha _i = \displaystyle\sum _{i=1} ^{4} \alpha _i^2 = \displaystyle\sum _{i=1} ^{4} \alpha _i^3 = \displaystyle\sum _{i=1} ^{4} \alpha _i^4 = -1$ , the expression becomes $16 \times 4 + (-1)(8 + 4 + 2 + 1) = 64- 15 = \boxed{49}$

I have solved this by calculus, but instead i will be posting a different solution.

Since $\alpha_1 ,\alpha_2 ,\alpha_3 ,\alpha_4$ are four of the five $5^{th}$ roots of unity, we have:

$\large{P(x)=x^{4}+x^{3}+x^{2}+x+1}$ which have the roots: $\alpha_1 ,\alpha_2 ,\alpha_3 ,\alpha_4$

Note that we need to find $\large{31.\sum{\frac{1}{2-x_k}}}$ where $x_k$ are the roots of $P(x)$ .

Let: $\large{y_k=\frac{1}{2-x_k}}$ so, we need to find $\sum{y_k}$

A short computation yields $\large{x_k=\frac{2y_k-1}{y_k}}$

Plugging this in $P(x)$ , we get:

$\large{(\frac{2y_k-1}{y_k})^{4}+(\frac{2y_k-1}{y_k})^{3}+(\frac{2y_k-1}{y_k})^{2}+\frac{2y_k-1}{y_k}+1=0}$

Multiplying by $y_k^{4}$ , we get

$\large{y_k^{4}+y_k^{3}(2y_k-1)+y_k^{2}(2y_k-1)^{2}+y_k(2y_k-1)^{3}+(2y_k-1)^{4}}$

It is easy to observe that :

$\text{coefficeint of}$ $y_k^{4}=1+2+2^{2}+2^{3}+2^{4}=31$

$\text{coefficeint of}$ $y_k^{3}=-(1+2.^{2}C_1+2^{2}.^{3}C_2+2^{3}.^{4}C_3)=-49$

$\large{\therefore \sum{y_k}=-\frac{-49}{31}=\frac{49}{31}}$

$\implies \large{31.\sum{\frac{1}{2-x_k}}=31.\sum{y_k}=\boxed{49}}$

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@Calvin Lin . i think you were thinking of something like this? :D

First of all, this method is long and not advisable too but just I am sharing for clarity.

Well, we can see that 5th root of unity means ${\alpha}^{5} = 1$

${\alpha}^{5} = {e}^{2n\iota\pi}$

$\alpha = {e}^{\frac{2n\iota\pi}{5}}$

So, ${\alpha}_{1} = {e}^{\frac{2\iota\pi}{5}}$

${\alpha}_{2} = {e}^{\frac{4\iota\pi}{5}}$

${\alpha}_{3} = {e}^{\frac{6\iota\pi}{5}}$

${\alpha}_{4} = {e}^{\frac{8\iota\pi}{5}}$

${\alpha}_{5} = {e}^{\frac{10\iota\pi}{5}} = {e}^{2\iota\pi}= 1$

And similarly, ${\alpha}_{6} = {e}^{\frac{12\iota\pi}{5}}$ , but wait,

${e}^{\frac{12\iota\pi}{5}} = {e}^{2\iota\pi + \frac{2\iota\pi}{5}} = {e}^{\frac{2\iota\pi}{5}}$

So, we got our 4 $5th$ roots of unity. Hence, substituting,

$31(\frac{1}{2- {e}^{\frac{2\iota\pi}{5}}} + \frac{1}{2- {e}^{\frac{4\iota\pi}{5}}} + \frac{1}{2- {e}^{\frac{6\iota\pi}{5}}} + \frac{1}{2- {e}^{\frac{8\iota\pi}{5}}})$

We know that ${e}^{\iota x} = cos x + \iota sin x$

Now, using some trigonometry, we can further simplify it as -

$31(1 + \frac{15 - 6(cos\frac{4\pi}{5} + cos\frac{2\pi}{5})}{25 - 20(cos\frac{4\pi}{5} + cos\frac{2\pi}{5}) + 16cos\frac{2\pi}{5}cos\frac{4\pi}{5}}$

We know that -

$cos\frac{4\pi}{5} + cos\frac{2\pi}{5} = -\frac{1}{2}$

$cos\frac{2\pi}{5}cos\frac{4\pi}{5} = -\frac{1}{4}$

Substituting both these in the above equation -

$31(1 + \frac{18}{31}) = \boxed{49}$

I hope this method to be the easiest..

The question involves Complex Numbers. Then, why can't we use complex guys?

Notice one thing

Let a general solution of the roots be $\alpha$

Fifth root of unity then can be

$\displaystyle\begin{array}{l|c|r} Given&1 &\alpha_1&\alpha_2&\alpha_3&\alpha_4\\ \hline Rename&1&\alpha&\alpha^2&\alpha^3&\alpha^4\end{array}$

And they're cyclic (or whatever term it is called)

Also notice

$\displaystyle\sum_{i=1}^4 \alpha_i = \sum \alpha = \sum \alpha^2= \sum \alpha^3= \sum \alpha^4\\$

Now begins the FIGHT

$\\\displaystyle 31 = 2^5 - \alpha^5\\ \Rightarrow\begin{aligned}\frac{31}{2-\alpha} &=\frac{2^5-\alpha^5}{2 - \alpha}\\&=2^4 + 2^3 \alpha+2^2\alpha^2+2 \alpha^3+\alpha^4\\\end{aligned}\\$

$\displaystyle\begin{aligned}\Rightarrow \sum \limits_1^4 \frac{2^5 - \alpha^5}{2-\alpha} &= 4\cdot 2^4 + 2^3\left(\sum \alpha\right) + 2^2\left(\sum \alpha^2\right) \nonumber \\ &\qquad{} +2\left(\sum \alpha^3\right)+\left(\sum \alpha^4 \right)\end{aligned}\\$

To get $\displaystyle \sum\alpha$ , take the equation

$x^5=1$ OR $x^5-1=0$

$\begin{aligned}\Rightarrow 1+\sum\alpha &= 0\\\Rightarrow \sum\alpha &= -1 \\\Rightarrow\sum \alpha &= \sum \alpha^2= \sum \alpha^3= \sum \alpha^4=-1\end{aligned}$

Hence the answer,

$\displaystyle\begin{aligned}\Rightarrow \sum \limits_1^4 \frac{2^5 - \alpha^5}{2-\alpha} &= 4\cdot2^4 - 2^3 - 2^2-2-1\\&=64-8-4-2-1\\&=49\end{aligned}$

$\huge\therefore \text{ Answer is }\boxed{49}$

Thanks for the nice question.

Let $n=\dfrac{31}{2-a}$

Some algebra yields $an=2n-31$

Raising both sides of this equation by 5, we get

$n^5=32n^5-5*2^431n^4+...$

By Vieta's , the sum of the five values of $n$ is $\dfrac{5*2^4*31}{31}=80$

But why are there five values of n when there are only four terms in our problem? This is because we included the non-existent term $n=\dfrac{31}{2-1}=31$ when we raised the equation to the 5th power.

Thus, the sum of the four terms in the problem is

$80-31=\boxed{49}$

Write a solution.