Discussion
\(2 + {}^{15}C_{1} + {}^{15}C_{2} + \ldots {}^{15}C_{14} =\)
Answer: B
The sum is \(2^{15}\).
The sum of all binomial coefficients of order 15 is \(2^{15}\):
\[ \begin{aligned} &{}^{15}C_{0} + {}^{15}C_{1} + \cdots + {}^{15}C_{15} \\ &= 2^{15} \end{aligned} \]
The given expression is missing only \({}^{15}C_{0}\) and \({}^{15}C_{15}\), each equal to 1, and the leading 2 supplies exactly these. So it equals the full sum, \(2^{15}\). \(2^{14}\) would be the sum of the even-index (or odd-index) coefficients alone, and \(2^{16}\) belongs to order 16.
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