Suppose P(x)=1−x+x2−x3+…+x2012 is expressed as a polynomial in y, as Q(y)=a0+a1y+…+a2012y2012 where y=x−2, then ∑i=02012 ai equal

Suppose $P (x) = 1 - x + x^{2} - x^{3} + \dots + x^{2012}$ is expressed as a polynomial in
$y,$ as $Q (y) = a_{0} + a_{1} y + \dots + a_{2012} y^{2012}$ where $y = x - 2,$ then $\sum_{i = 0}^{2012} a_{i}$ equal

We have

$P (x) = \frac{1 - (- x)^{2013}}{1 - (- x)} = \frac{1 + x^{2013}}{1 + x}$

$\sum_{i = 0}^{2012} a_{i} = Q (1) = P (3) [∵ y = 1 = x - 2]$

$= \frac{1}{4} (3^{2013} + 1)$

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