Mathematics - Probability Question with Solution | TestHub

Divisors of $N=2^{n-1}(2^n-1)$ are:

$$\begin{array}{rl}& 1,2,2^2,\dots ,2^{n-1},2^n-1,2(2^n-1),2^2(2^n-1),\dots ,2^{n-1}(2^n-1)\\ & 1+\frac{1}{{\alpha }_1}+\frac{1}{{\alpha }_2}+\dots +\frac{1}{{\alpha }_k}=1+\frac{1}{2}+\dots +\frac{1}{2^{n-1}}+\frac{1}{2^n-1}+\frac{1}{2(2^n-1)}+\dots +\frac{1}{2^{n-1}(2^n-1)}\\ & =\left(1+\frac{1}{2}+\dots +\frac{1}{2^{n-1}}\right)+\frac{1}{2^n-1}\left(1+\frac{1}{2}+\dots +\frac{1}{2^{n-1}}\right)\\ & =\left(1+\frac{1}{2}+\dots +\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^n-1}\right)\\ & =\frac{1-{\left(\frac{1}{2}\right)}^n}{1-\frac{1}{2}}\left(\frac{2^n}{2^n-1}\right)=\frac{2^n-1}{2^{n-1}}\left(\frac{2^n}{2^n-1}\right)=2\end{array}$$