Question

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

Answer 1

$A =\displaystyle\sum_{k=1}^{a}{ }^{n} C_{k}\left[\left(-\frac{1}{2}\right)^{k}+\left(\frac{-3}{4}\right)^{k}+\left(\frac{-7}{8}\right)^{k}+\left(\frac{-15}{16}\right)^{k}+\left(\frac{-37}{32}\right)^{k}\right]$
$A =\left(1-\frac{1}{2}\right)^{ n }+\left(1-\frac{3}{4}\right)^{ n }+\left(1-\frac{7}{8}\right)^{ n }+\left(1-\frac{15}{16}\right)^{ n }+\left(1-\frac{31}{32}\right)^{ a }$
$A =\frac{1}{2^{ n }}+\frac{1}{4^{ n }}+\frac{1}{8^{ n }}+\frac{1}{16^{ n }}+\frac{1}{32^{ n }}$
$A=\frac{1}{2^{n}}\left(\frac{1-\left(\frac{1}{2^{n}}\right)^{5}}{1-\frac{1}{2^{n}}}\right) \Rightarrow A=\frac{\left(1-\frac{1}{2^{5 n}}\right)}{\left(2^{n}-1\right)}$
$\left(2^{n}-1\right) A =1-\frac{1}{2^{5 n }}$, Given $63 A =1-\frac{1}{2^{30}}$
Clearly $5 n =30$
$n =6$