Simply Linear

Algebra Level 5

n = 0 2015 ( 1 ) ( n + 1 2 ) ( x + n ) 3 = 0 \sum_{n=0}^{2015} (-1)^{\left\lfloor\left(\frac{n+1}{2}\right)\right\rfloor} \cdot (x + n)^3 = 0

Let α \alpha be the root of the above equation. Find 2 α + 2016 2\alpha + 2016 .

Notations

  • Floor function is the exponent of (-1).
  • \lfloor \cdot \rfloor denotes the floor function .


The answer is 1.

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1 solution

Rajen Kapur
Jan 9, 2016

The given function can be split into 504 chunks each of 4 terms ( x + k + 3 ) 3 ( x + k + 2 ) 3 ( x + k + 1 ) 3 + ( x + k ) 3 (x + k + 3)^3 - (x + k + 2)^3 - (x + k + 1)^3 + (x + k)^3 that get simplified to 12 ( x + k ) + 18 12(x + k) +18 , where k = 0, 4, 8, . . . . , 2012 thus transforming the expression into a linear equation giving α = 2015 2 \alpha = \frac{-2015}{2} .

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