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 $63A=1-\left(\frac{1}{2^{30}}\right)$, then $n$ is equal to

Answer 1

Finding the value of $n$:

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

We know that, $\sum _{k=0}^n{}^{n}C_k{x}^k={\left(1+x\right)}^n$

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

This is a G.P series whose sum of terms is given as below,

$$\begin{gathered} S=\frac{a\left(1-r^n\right)}{1-r}\left[r=\frac{1}{2^n},a=\frac{1}{2^n},n=5\right] \\ \Rightarrow S=\frac{{\displaystyle \frac{1}{2^n}}\left(1-{\left({\displaystyle \frac{1}{2^n}}\right)}^5\right)}{1-{\displaystyle \frac{1}{2^n}}} \\ \Rightarrow S=\frac{{\displaystyle \frac{1}{2^n}}\left(1-{\displaystyle \frac{1}{2^{5n}}}\right)}{{\displaystyle \frac{2^n-1}{2^n}}} \\ \Rightarrow S=\frac{1-{\displaystyle \frac{1}{2^{5n}}}}{2^n-1} \end{gathered}$$

Now it is also given that, $63A=1-\left(\frac{1}{2^{30}}\right)$ Therefore,

$$\begin{gathered} 63A=1-\frac{1}{2^{30}} \\ \Rightarrow 63\left(\frac{1-{\displaystyle \frac{1}{2^{5n}}}}{2^n-1}\right)=1-\frac{1}{2^{30}} \\ \Rightarrow 1-\frac{1}{2^{5n}}=\left(1-\frac{1}{2^{30}}\right)\times \left(\frac{2^n-1}{63}\right) \end{gathered}$$

By observing the above expression we get,

$$\begin{gathered} 5n=30 \\ \Rightarrow n=6 \end{gathered}$$

Hence, the value of $n$is equal to $6$.