If (1 - x + x<sup>2</sup>)<sup>n</sup> = a<sub>0</sub> + a<sub>1</sub>x + a<sub>2</sub>x<sup>2</sup> + ... + a<sub>2n</sub>x<sup>2n</sup>, then a<sub>0</sub> + a<sub>2</sub> + a<sub>4</sub> + ... a<sub>2n</sub> is?
Step-by-step Solution:
We have, \[ a_1 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + \dots + a_{2n}x^{2n} = (1 - x + x^2)^n \] On putting \( x = 1 \) and \( x = -1 \) respectively, we get \[ (a_0 + a_2 + a_4 + \dots) + (a_1 + a_3 + a_5 + \dots) = 1 \quad \text{..........(i)} \] \[ (a_0 + a_2 + a_4 + \dots) - (a_1 + a_3 + a_5 + \dots) = 3^n \quad \text{..........(ii)} \] On adding Eqs. (i) and (ii), we get \[ 2(a_0 + a_2 + a_4 + \dots) = 3^n + 1 \] \[ a_0 + a_2 + a_4 + \dots = \frac{3^n + 1}{2} \]