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Q.If (1+x+x2)20=∑r=040arxr(1 + x + x^2)^{20} = \sum_{r=0}^{40} a_r x^r, then a0+a3+a6+⋯+a39a_0 + a_3 + a_6 + \cdots + a_{39} equals

a
320+ω20+ω403\dfrac{3^{20} + \omega^{20} + \omega^{40}}{3}
b
3193^{19}
c
320+13\dfrac{3^{20} + 1}{3}
d
320/33^{20}/3

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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