Discussion

If \(\sum_{k=0}^{n+1} {}^{n+1}C_{k} = 512\), then \(\sum_{k=0}^{n} {}^{n}C_{k} =\)

Answer: B

The second sum is 256.

The binomial coefficients of a power add to a power of 2: \(\sum_{k=0}^{m} {}^{m}C_{k} = 2^{m}\). So

\[ \begin{aligned} 2^{n+1} &= 512 = 2^{9} \;\Rightarrow\; n = 8 \\ \sum_{k=0}^{8} {}^{8}C_{k} &= 2^{8} = 256 \end{aligned} \]

Each step down in \(n\) halves the sum. 512 repeats the given value, 128 halves it twice, and 511 or 510 are not powers of 2.

Here \(n = 8\) itself is not offered as an option; the question asks for the sum, \(2^{8}\).

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