Let n be a positive integer. Let A = ∑ k = 0 n - 1 k × C k n 1 2 k + 3 4 k + 7 8 k + 15 16 k + 31…

$A=\sum _{k=0}^n{}^{n}C_k\left[{\left(-\frac{1}{2}\right)}^k+{\left(\frac{-3}{4}\right)}^k+{\left(\frac{-7}{8}\right)}^k+{\left(\frac{-15}{16}\right)}^k+{\left(\frac{-31}{32}\right)}^k\right]$

$A={\left(1-\frac{1}{2}\right)}^n+{\left(1-\frac{3}{4}\right)}^n+{\left(1-\frac{7}{8}\right)}^n+{\left(1-\frac{15}{16}\right)}^n+{\left(1-\frac{31}{32}\right)}^n$

$A=\frac{1}{2^n}+\frac{1}{4^n}+\frac{1}{8^n}+\frac{1}{{16}^n}+\frac{1}{{32}^n}$

$A=\frac{1}{2^n}\left(\frac{1-{\left(\frac{1}{2^n}\right)}^5}{1-\frac{1}{2^n}}\right)\Rightarrow A=\frac{\left(1-\frac{1}{2^{5n}}\right)}{\left(2^n-1\right)}$

$\left(2^n-1\right)A=1-\frac{1}{2^{5n}},$

Given, $63A=1-\frac{1}{2^{30}}$

Clearly $5n=30$

$n=6$