Let n be a positive integer. Let A = sum_{k=0 | vedclass

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(C) We know that $\sum_{k=0}^{n} {}^{n}C_{k} x^{k} = (1+x)^{n}$.Given $A = \sum_{k=0}^{n} {}^{n}C_{k} \left[ (-\frac{1}{2})^{k} + (-\frac{3}{4})^{k} +

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$6$

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(C) We know that $\sum_{k=0}^{n} {}^{n}C_{k} x^{k} = (1+x)^{n}$.Given $A = \sum_{k=0}^{n} {}^{n}C_{k} \left[ (-\frac{1}{2})^{k} + (-\frac{3}{4})^{k} + (-\frac{7}{8})^{k} + (-\frac{15}{16})^{k} + (-\frac{31}{32})^{k} \right]$.Using the binomial expansion,we get:$A = (1 - \frac{1}{2})^{n} + (1 - \frac{3}{4})^{n} + (1 - \frac{7}{8})^{n} + (1 - \frac{15}{16})^{n} + (1 - \frac{31}{32})^{n}$.$A = (\frac{1}{2})^{n} + (\frac{1}{4})^{n} + (\frac{1}{8})^{n} + (\frac{1}{16})^{n} + (\frac{1}{32})^{n}$.$A = \frac{1}{2^{n}} + \frac{1}{2^{2n}} + \frac{1}{2^{3n}} + \frac{1}{2^{4n}} + \frac{1}{2^{5n}}$.This is a geometric progression with $a = \frac{1}{2^{n}}$ and $r = \frac{1}{2^{n}}$ for $5$ terms.$A = \frac{1}{2^{n}} \left( \frac{1 - (\frac{1}{2^{n}})^{5}}{1 - \frac{1}{2^{n}}} \right) = \frac{1}{2^{n}} \left( \frac{1 - \frac{1}{2^{5n}}}{\frac{2^{n}-1}{2^{n}}} \right) = \frac{1 - \frac{1}{2^{5n}}}{2^{n}-1}$.Thus,$(2^{n}-1)A = 1 - \frac{1}{2^{5n}}$.Given $63A = 1 - \frac{1}{2^{30}}$,we compare $2^{n}-1 = 63$ and $5n = 30$.$2^{n} = 64 \Rightarrow n = 6$ and $5n = 30 \Rightarrow n = 6$.Therefore,$n = 6$.

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