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Q.If (1+x)20=∑k=020(20k)xk(1 + x)^{20} = \sum_{k=0}^{20} \binom{20}{k} x^k, then ∑k=010(202k)\sum_{k=0}^{10} \binom{20}{2k} equals

a
2192^{19}
b
219+(2010)2^{19} + \binom{20}{10}
c
219−(2010)2^{19} - \binom{20}{10}
d
2202^{20}

Correct Answer: Option A

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

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