When will the holidays come 1

We can derive the formula $\sum_{w}\frac{1}{a-w}=\frac{na^{n-1}}{a^n-1}$ for the $n$ th roots of unity $w$ , where $a$ fails to be an $n$ th root of unity. Let $n=5$ and $a=2$ to obtain the required value $\boxed{18}$ (subtract 1 since the problem starts the count at $k=1$ )

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To derive the formula, we make the substitution $x=\frac{1}{a-w}$ or $w=\frac{ax-1}{x}$ . Then $w^n=1$ gives $(ax-1)^n=x^n$ or $(a^n-1)x^n-na^{n-1}x^{n-1}+...=0$ , and our formula follows from Viete's rule, by finding the sum of roots of the transformed equation.

We have that, $f(x)=x^5-1=(x-1)(\prod_{i=1}^{4}(x-\alpha_{i}))\\ \Longrightarrow \ln(f(x))=\ln(x-1)+\sum_{i=1}^{4}(\ln(x-\alpha_{i}))\\ \Longrightarrow \dfrac{d}{dx}\ln(f(x))=\dfrac{1}{x-1}+\sum_{i=1}^{4}(\dfrac{1}{x-\alpha_{i}})\\ \Longrightarrow \dfrac{1}{f(x)} \times f^{'}(x)=\dfrac{1}{x-1}+\sum_{i=1}^{4}(\dfrac{1}{x-\alpha_{i}})$ Now,just put $x=2$ , $\dfrac{1}{f(2)}\times f^{'}(2)=1+\text{Required expression}\\ f(2)=31,f^{'}(x)=5x^4=80\\ \Longrightarrow \text{Required expression}=\dfrac{80}{31}-1=\dfrac{49}{31}\\ \Longrightarrow \text{Answer}=18$ And done!

$Let~~\theta=\dfrac {2\pi} 5.~~~\therefore~~\displaystyle \alpha_n={\huge e}^{i*\dfrac{2\pi} 5*n}=Cos(\theta*n)+iSin(\theta*n)...n=1, 2, 3, 4\\ Cos(\theta *4)=Cos(-\theta*4)=Cos(\theta*1) =\dfrac{\sqrt5-1} 4.\\ Cos(\theta*3)=Cos(-\theta*3)=Cos(\theta*2) =\color{#3D99F6}{-}\dfrac{\sqrt5+1} 4.\\ Sin(\theta*1) + Sin(\theta*4)=0. ~~~~~Sin(\theta*2) + Sin(\theta*3)=0. \\ \dfrac 1 {2-\alpha_n}= \dfrac 1 {2-\Big (Cos(\theta*n)+iSin(\theta*n)\Big )}~~~Rationalizing,\\ = \dfrac {(2-Cos(\theta*n))-iSin(\theta*n)}{(2-Cos(\theta*n))^2-(iSin\theta*n)^2 }= \dfrac {(2-Cos(\theta*n))-iSin(\theta*n)}{5-4*Cos(\theta*n) } \\ \therefore~ \dfrac 1 {2-\alpha_1} = \dfrac 1 {2-(Cos(\theta*1)+iSin(\theta*1))}= \dfrac {(2-\dfrac{\sqrt5-1} 4) - iSin(\theta*1)}{5-4*\dfrac{\sqrt5-1} 4} \\ = \dfrac {\dfrac{ 9 -\sqrt5} 4-iSin(\theta*1)}{6-\sqrt5} \\ \therefore~\dfrac 1 {2-\alpha_4} = \dfrac 1 {2-(Cos(\theta*4)+iSin(\theta*4))}= \dfrac {(2-\dfrac{\sqrt5-1} 4)-iSin(\theta*4)}{5-4*\dfrac{\sqrt5-1} 4 } \\ = \dfrac { \dfrac{9 - \sqrt5 } 4-iSin(\theta*4)}{6-\sqrt5} \\ \therefore~~ \dfrac 1 {2-\alpha_1} + \dfrac 1 {2-\alpha_4}=\dfrac { \dfrac{ 9 -\sqrt5} 4~~{\color{#3D99F6}{-iSin(\theta*1)} }+\dfrac{ 9 -\sqrt5} 4~~{\color{#3D99F6}{-iSin(\theta*4)}}} {6-\sqrt5} \\ =\frac 1 2 *\Big \{\dfrac{9-\sqrt5}{6-\sqrt5} \Big \} \\ Similarly~~ \dfrac 1 {2-\alpha_2} + \dfrac 1 {2-\alpha_3}= 2 *\dfrac{2-(-\dfrac{\sqrt5+1} 4)}{5-\{-(\sqrt5+1)\}}\\ =\frac 1 2 *\dfrac{9+\sqrt5}{6+\sqrt5}.\\ \therefore~\displaystyle \sum_{n=1}^4 \dfrac 1 {2- \alpha_n}=\frac 1 2 *\Big \{ \dfrac{9-\sqrt5}{6-\sqrt5}+\dfrac{9+\sqrt5}{6+\sqrt5} \Big \}\\ =\frac 1 2 *\Big \{ \dfrac{54-5-3\sqrt5+54-5+3\sqrt5}{36 -5} \Big \} =\dfrac{49}{31}=\dfrac a b .\\ a-b=\huge ~~\color{#D61F06}{18}.$

Let $f\left( x \right) ={ x }^{ 5 }-1=\left( x-1 \right) \prod _{ i=1 }^{ 4 }{ \left( x-{ a }_{ i } \right) }$

Also,

$g\left( x \right) =f\left( x+2 \right) ={ (x+2) }^{ 5 }-1=\left( x-(1-2) \right) \prod _{ i=1 }^{ 4 }{ \left( x-({ a }_{ i }-2) \right) }$

Therefore,

$g\left( \frac { 1 }{ x } \right) ={ (\frac { 1 }{ x } +2) }^{ 5 }-1=\left( \frac { 1 }{ x } -(1-2) \right) \prod _{ i=1 }^{ 4 }{ \left( \frac { 1 }{ x } -({ a }_{ i }-2) \right) }$

Now we can see roots of $g\left( \frac { 1 }{ x } \right)$ are $-1,\frac { 1 }{ { a }_{ i }-2 }$ for $i=1,2,3,4$ .

Also,

${(\frac{ 1 }{ x }+2) }^{ 5 }-1={ \left( \frac {1+2x}{x} \right) }^{ 5 }-1$ $=\frac { { \left( 1+2x \right) }^{ 5 }-{ x }^{ 5 } }{ { x }^{ 5 }\\ }$

Therefore, $g\left( \frac { 1 }{ x } \right) =0$ $=\frac { { \left( 1+2x \right) }^{ 5 }-{ x }^{ 5 } }{ { x }^{ 5 }\\ }$

${ \Rightarrow \left( 1+2x \right) }^{ 5 }-{ x }^{ 5 }=0$

Therefore, $g\left( \frac { 1 }{ x } \right) =({ 2 }^{ 5 }-1){ x }^{ 5 }+5\times { 2 }^{ 4 }{ x }^{ 4 }+$ something

So sum of roots of $g\left( \frac { 1 }{ x } \right)$ is $-1+\sum _{ i=1 }^{ 4 }{ \frac { 1 }{ { a }_{ i }-2 } }=\frac { -5\times { 2 }^{ 4 } }{ ({ 2 }^{ 5 }-1) } =\frac { -80 }{ 31 }$

$\Rightarrow \sum _{ i=1 }^{ 4 }{ \frac { 1 }{ { a }_{ i }-2 } } =\frac { -49 }{ 31 } \Rightarrow \sum _{ i=1 }^{ 4 }{ \frac { 1 }{ 2-{ a }_{ i } } } =\frac { 49 }{ 31 }$

Therefore, $a-b=18$ .

All the roots are of the equation x5 – 1 = 0 Σ α = Σ αβ= Σ α βγ =Σ α β γδ= 0 Πα = 1

Let α5 = 1 so, Σ {1/(2- α i)} numerator is, 24 – (α2 + α3 + α4 + α5) 23 +…….. = 5.24 –(α1+ α2 + α3 + α4 + α5) 4.23 +…….. = 5.24 = 80 denominator is, Π(2 - α i) = 2^5 – 1 = 31 as Π (x - α i)=x^5 – 1

so a/b = (80/31) – 1/(2-1) = 49/31 a-b = 49 – 31 = 18

In the original sum was taken x instead of two. Let x be such that x^5-1=0. The sum requested is equal to (4x^3+3x^2+2x+1)(x-1)/(x^5 - 1) with x different from one. For " x = 2", we get the sum equal to 49/31, so a-b=18.

If we note that $\frac{1}{2}|\alpha _{i}|=\frac{1}{2}$ then we can write

$\frac{1}{2-\alpha _{i}}=\frac{1}{2} (1+\frac{\alpha _{i}}{2}+(\frac{\alpha _{i}}{2})^{2}+... )$

If we sum the above equation 'column-wise' from $i=1$ to $i=5$ and use that $1+\alpha _{1}+...+\alpha _{4}=0$ , and that $(\alpha _{1})^{2}=\alpha _{2}$ , $(\alpha _{1})^{3}=\alpha _{3}$ , etc. then

$\frac{1}{2-1}+\sum_{i=1}^4 \frac{1}{2-\alpha _{i}}=\frac{1}{2} (1+\frac{5}{2^{5}}+\frac{5}{2^{10}}+...)$

The RHS is a geometric series and hence by summing it and subtracting 1 from both sides we get our desired sum is $\boxed{\frac{49}{31}}$