\(a_n<a_{n+1}\) for all positive integers \(n\),
and a line passing through points
\(\mathrm{P}_n\big(a_n,2^{a_n}\big)\) and \(\mathrm{P}_{n+1}\big(a_{n+1},2^{a_{n+1}}\big)\)
has a slope of \(k\times 2^{a_n}\).
A line passing through points \(\mathrm{P}_n\) and \(\mathrm{P}_{n+1}\)
has a slope of \(k\times 2^{a_n}\), so
\(2^{a_{n+1}-a_n}=k(a_{n+1}-a_n)+1\).
In other words, \(a_{n+1}-a_n\) is a solution to the equation
\(2^x=kx+1\) for all positive integers \(n\).
Since \(k>1\), the equation \(2^x=kx+1\) only has one positive solution \(d\).
Therefore \(a_{n+1}-a_n=d\) for all \(n\),
so \(\{a_n\}\) is an arithmetic progression with a common difference of \(d\).
Since point \(\mathrm{Q_n}\) has coordinates \(\big(a_n+1,2^{a_n}\big)\),
\(A_n=\dfrac{1}{2}(a_{n+1}-a_n)\big(2^{a_{n+1}}-2^{a_n}\big)\).
Since \(\dfrac{A_3}{A_1}=16\), \(d=\fbox{\(\;(\alpha)\;\)}\),
and the general term of the sequence \(\{a_n\}\) is
\(a_n=\fbox{\(\;(\beta)\;\)}\).
Therefore \(A_n=\fbox{\(\;(\gamma)\;\)}\) for all positive integers \(n\).
Let \(p\), be the correct number for \((\alpha)\),
and let \(f(n)\) and \(g(n)\) be the correct expression for
\((\beta)\) and \((\gamma)\) respectively.
What is the value of \(p+\dfrac{g(4)}{f(2)}\)?