Discussion
\({}^{21}C_{1} + {}^{21}C_{2} + ... + {}^{21}C_{10} =\)
Answer: E
The sum is \(2^{20} - 1\).
The 22 coefficients of order 21 add to \(2^{21}\), and they pair up by symmetry: \({}^{21}C_{r} = {}^{21}C_{21-r}\). So \({}^{21}C_{0}\) to \({}^{21}C_{10}\) make up exactly half.
\[ \begin{aligned} \sum_{r=0}^{10} {}^{21}C_{r} &= \frac{2^{21}}{2} = 2^{20} \\ \sum_{r=1}^{10} {}^{21}C_{r} &= 2^{20} - 1 \end{aligned} \]
The sum starts at r = 1, so the term \({}^{21}C_{0} = 1\) is removed. \(2^{20}\) forgets that, and \(2^{21} - 1\) or \(2^{21} - 2\) treat the sum as if it ran over all the coefficients.
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