Question

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

Answer 1

Answer: 6

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