Aspire Faculty ID #16411 · Topic: NIMCET 2009 · Just now
NIMCET 2009

If $(1 + x - 2x^2)^6 = 1 + a_1 x + a_2 x^2 + \ldots + a_{12} x^{12}$, then the value of $a_2 + a_4 + a_6 + \ldots + a_{12}$ is:

Solution

Even–power coefficient sum = $\dfrac{f(1) + f(-1)}{2}$ 
 $f(1) = (1 + 1 - 2)^6 = 0^6 = 0$ 
$f(-1) = (1 - 1 - 2)^6 = (-2)^6 = 64$ 
 Required sum $= \dfrac{0 + 64}{2} = 32$

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