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Algebra Level 5

∑ k = 1 10 w k 2 − w k \large \displaystyle \sum _{k=1}^{10}\dfrac{w^k}{2-w^k}

Let w = e 2 i π / 10 w=e^{2i\pi/10} . If the value of the summation above is equal to a b \dfrac ab , where a a and b b are coprime positive integers, find a + b a+b .


The answer is 1033.

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Julian Poon by Julian Poon , 20, Singapore

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2 solutions

Otto Bretscher
Apr 12, 2016

Substitute z = x 2 − x z=\frac{x}{2-x} or x = 2 z 1 + z x=\frac{2z}{1+z} . Now x n = 1 x^n=1 gives ( 2 z ) n = ( 1 + z ) n (2z)^n=(1+z)^n or ( 2 n − 1 ) z n − n z n − 1 + . . . = 0. (2^n-1)z^n-nz^{n-1}+...=0. By Viète, the sum of the roots is n 2 n − 1 \frac{n}{2^n-1} . For n = 10 n=10 this is 10 1023 \frac{10}{1023} , and the answer is 1033 \boxed{1033}

14 Helpful 1 Interesting 1 Brilliant 0 Confused

Let α = e 2 π i 10 \alpha = e^{\frac{2πi}{10}}
S = ∑ k = 1 10 α k x − α k S = \displaystyle \sum_{k=1}^{10}\dfrac{\alpha^{k}}{x-\alpha^{k}}

S = − 10 + ∑ k = 1 10 x x − α k S = -10 + \displaystyle \sum_{k=1}^{10} \dfrac{x}{x-\alpha^{k}}
( x − α ) ⋅ ( x − α 2 ) … ( x − α 10 ) = x 10 − 1 (x-\alpha)\cdot(x-\alpha^{2})\ldots(x-\alpha^{10}) = x^{10}-1
Taking logarithm and differntiating,
∑ k = 1 10 1 x − α k = 10 x 9 x 10 − 1 \displaystyle \sum_{k=1}^{10}\dfrac{1}{x-\alpha^{k}} = \dfrac{10x^{9}}{x^{10}-1}


∴ S = − 10 + 10 x 10 x 10 − 1 \therefore S = -10 + \dfrac{10x^{10}}{x^{10}-1}
∴ S = 10 x 10 − 1 \therefore S = \dfrac{10}{x^{10}-1}
Put x = 2,
S = 10 1023 S = \dfrac{10}{1023}
Answer : 1033 \text{Answer :} 1033

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