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Q.If (1+x)15=∑r=015(15r)xr(1 + x)^{15} = \sum_{r=0}^{15} \binom{15}{r} x^r, then ∑r=07(152r+1)\sum_{r=0}^{7} \binom{15}{2r+1} equals

a
2142^{14}
b
215−(158)2^{15} - \binom{15}{8}
c
214+(158)2^{14} + \binom{15}{8}
d
2152^{15}

Correct Answer: Option A

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

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