\(a_{n+1}=S_{n+1}-S_n\) for all positive integers \(n\),
so the given equation becomes
\(S_{n+1}=(n+2)S_n+n!\quad(n\geq 1)\).
Dividing both sides with \((n+2)!\) gives
\(\dfrac{S_{n+1}}{(n+2)!}=\dfrac{S_n}{(n+1)!}+\dfrac{1}{(n+1)(n+2)}\).
Let \(b_n=\dfrac{S_n}{(n+1)!}\).
Then \(b_1=\dfrac{1}{2}\), and
\(b_{n+1}=b_n+ \dfrac{1}{(n+1)(n+2)}\).
The general term of the sequence \(\{b_n\}\) is
\(b_n=\dfrac{\fbox{\(\;(\alpha)\;\)}}{n+1}\).
Therefore
\(S_n=\fbox{\(\;(\alpha)\;\)}\times n!\)
and
\(a_n=\fbox{\(\;(\beta)\;\)}\times (n-1)!\quad(n\geq1)\).
Let \(f(n)\) and \(g(n)\) be the correct expression
for \((\alpha)\) and \((\beta)\) respectively.
What is the value of \(f(7)+g(6)\)?