Another form presents geometric sequence

Algebra Level 3

k = 1 2016 x k = x 2016 1 x \large \sum_{k=1}^{2016}x^k=\frac{x-2016}{1-x}

If the real root of the equation above is of the form b a \large \sqrt[a]{b} , where a a and b b are positive integers with a a minimized, find a + b a+b .


The answer is 4033.

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1 solution

Tommy Li
Aug 16, 2016

k = 1 2016 x k = x 2016 1 x \large \displaystyle\sum_{k=1}^{2016}x^k=\frac{x-2016}{1-x}

x ( x 2016 1 ) x 1 = x 2016 1 x \large \dfrac{x(x^{2016}-1)}{x-1}=\dfrac{x-2016}{1-x}

x 2017 x x 1 = 2016 x x 1 \large \dfrac{x^{2017}-x}{x-1}=\dfrac{2016-x}{x-1}

x 2017 x = 2016 x \large x^{2017}-x=2016-x

x 2017 = 2016 \large x^{2017}=2016

x = 2016 2017 \large x=\sqrt[2017]{2016}

a + b = 2017 + 2016 = 4033 \large \Rightarrow a+b = 2017+2016 = 4033

Not 2016 this time!

Alex G - 4 years, 10 months ago

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