Discussion
\(1 + {}^{100}C_{1} + {}^{100}C_{2} + ... + {}^{100}C_{99} + 1 =\)
Answer: D
The sum is \(2^{100}\).
The first 1 is \({}^{100}C_{0}\) and the last 1 is \({}^{100}C_{100}\), so the expression is the full sum of binomial coefficients of order 100.
\[ \sum_{r=0}^{100} {}^{100}C_{r} = (1 + 1)^{100} = 2^{100} \]
\(2^{99}\) is half that sum (the even or the odd coefficients alone), and \(2^{101}\) doubles it. \(100^{2}\) has no connection with the binomial sum.
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