Question

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

Answer 1

$A=\sum _{k=0}^n(-1{)}^k{}^n{C}_k[{\left(\frac{1}{2}\right)}^k+{\left(\frac{3}{4}\right)}^k+{\left(\frac{7}{8}\right)}^k+{\left(\frac{15}{16}\right)}^k+{\left(\frac{31}{32}\right)}^k]$
$\Rightarrow A={(1-\frac{1}{2})}^n+{(1-\frac{3}{4})}^n+{(1-\frac{7}{8})}^n+{(1-\frac{15}{16})}^n+{(1-\frac{31}{32})}^n$
$\Rightarrow A={\left(\frac{1}{2}\right)}^n+{\left(\frac{1}{4}\right)}^n+{\left(\frac{1}{8}\right)}^n+{\left(\frac{1}{16}\right)}^n+{\left(\frac{1}{32}\right)}^n$
$=\frac{1}{2^n}+\frac{1}{2^{2n}}+\frac{1}{2^{3n}}+\frac{1}{2^{4n}}+\frac{1}{2^{5n}}$
$=\frac{1}{2^n}\left[\frac{1-{\left(\frac{1}{2^n}\right)}^5}{1-\frac{1}{2^n}}\right]$
$A=\frac{2^{5n}-1}{2^{5n}(2^n-1)}$

$63A=\frac{63(2^{5n}-1)}{2^{5n}(2^n-1)}$
$=\frac{63}{2^n-1}(1-\frac{1}{2^{5n}})=1-\frac{1}{2^{30}}$
$\Rightarrow n=6$