Discussion
Let \(\alpha = \sum_{k=0}^{5} {}^{10}C_{2k}\) and \(\beta = \sum_{k=0}^{4} {}^{10}C_{2k+1}\).Then \(\alpha - \beta\) is equal to
Answer: E
\(\alpha - \beta = \) 0.
\(\alpha\) is the sum of the even-indexed coefficients \({}^{10}C_{0}, {}^{10}C_{2}, \ldots, {}^{10}C_{10}\), and \(\beta\) the odd-indexed ones \({}^{10}C_{1}, \ldots, {}^{10}C_{9}\). Put \(x = -1\) in \((1 + x)^{10}\):
\[ \begin{aligned} (1 - 1)^{10} &= \sum_{r=0}^{10} (-1)^{r}\,{}^{10}C_{r} \\ 0 &= \alpha - \beta \end{aligned} \]
Each of \(\alpha\) and \(\beta\) separately equals \(2^{9} = 512\), half of \(2^{10}\). Options 32, 64, 128 and 256 confuse the difference with one of these sums.
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