# Let $n$ be a positive integer.

Question:

Let $n$ be a positive integer. Let

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

If $63 \mathrm{~A}=1-\frac{1}{2^{30}}$, then $\mathrm{n}$ is equal to_______.

Solution:

$A=\sum_{k=0}^{n}{ }^{n} C_{k}\left[\left(-\frac{1}{2}\right)^{k}+\left(\frac{-3}{4}\right)^{k}+\left(\frac{-7}{8}\right)^{k}+\left(\frac{-15}{16}\right)^{k}+\left(\frac{-37}{32}\right)^{k}\right]$

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

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

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

$\left(2^{\mathrm{n}}-1\right) \mathrm{A}=1-\frac{1}{2^{5 \mathrm{n}}}$, Given $63 \mathrm{~A}=1-\frac{1}{2^{30}}$

Clearly $5 \mathrm{n}=30$

$\mathrm{n}=6$