Let $\mathrm{n}$ be a positive integer. Let\
$\mathrm{A}=\sum_{\mathrm{k}=0}^{\mathrm{n}}(-1)^{\mathrm{k}} \mathrm{n}_{\mathrm{C}_{\mathrm{k}}}\left[\left(\frac{1}{2}\right)^{\mathrm{k}}+\left(\frac{3}{4}\right)^{\mathrm{k}}+\left(\frac{7}{8}\right)^{\mathrm{k}}+\left(\frac{15}{16}\right)^{\mathrm{k}}+\left(\frac{31}{32}\right)^{\mathrm{k}}\right]$
If $63 \mathrm{~A}=1-\frac{1}{2^{30}}$, then $\mathrm{n}$ is equal to
$\mathrm{A}=\sum_{k=0}^{n}{ }^{n} \mathrm{C}_{\mathrm{k}}\left[\left(-\frac{1}{2}\right)^{2}+\left(\frac{-3}{4}\right)^{2}+\left(\frac{-7}{8}\right)^{2}+\left(\frac{-15}{16}\right)^{2}+\left(\frac{-37}{32}\right)^{2}\right]_{\mathrm{n}}$
$A=\left(1-\frac{1}{2}\right)^{2}+\left(1-\frac{3}{4}\right)^{n}+\left(1-\frac{7}{8}\right)^{=}+\left(1-\frac{15}{16}\right)^{n}+\left(1-\frac{31}{32}\right)^{n}$
$\mathrm{A}=\frac{1}{2^{n}}+\frac{1}{4^{\mathrm{n}}}+\frac{1}{8^{n}}+\frac{1}{16^{n}}+\frac{1}{32^{\mathrm{n}}}$
$A=\frac{1}{2^{\mathrm{n}}}\left(\frac{1-\left(\frac{1}{2^{\mathrm{n}}}\right)^{5}}{1-\frac{1}{2^{\mathrm{n}}}}\right) \Rightarrow A=\frac{\left(1-\frac{1}{2^{5 \mathrm{n}}}\right)}{\left(2^{\mathrm{n}}-1\right)}$
$\left(2^{n}-1\right) \mathrm{A}=1-\frac{1}{2^{5 n}}$, Given $63 \mathrm{~A}=1-\frac{1}{2^{30}}$
Clearly $5 \mathrm{n}=30$
$\mathrm{n}=6$
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