Answer: (a) \(2!-S_{1}=1,\;3!-S_{2}=1,\;4!-S_{3}=1,\;5!-S_{4}=1\)
(b) \(S_{n}=(n+1)!-1\)
(c) Let \(H_n\) be the statement \(S_n=(n+1)!-1\).
For \(n=1\), \(S_1=u_1=1\cdot 1!=1\), so \(S_1=2!-1\). Hence \(H_1\) is true.
Assume \(H_k\) is true for some positive integer \(k\), so \(S_k=(k+1)!-1\).
Then
\(S_{k+1}=S_k+u_{k+1}=(k+1)!-1+(k+1)(k+1)!\).
So
\(S_{k+1}=(k+1)!\bigl(1+k+1\bigr)-1=(k+1)!(k+2)-1=(k+2)!-1\).
Thus \(H_{k+1}\) is true whenever \(H_k\) is true.
Therefore, by mathematical induction, \(S_n=(n+1)!-1\) for all positive integers \(n\).
Given \(u_r=r\times r!\), we have
\(u_1=1\cdot1!=1,\quad u_2=2\cdot2!=4,\quad u_3=3\cdot3!=18,\quad u_4=4\cdot4!=96\).
Hence
\(S_1=1\), so \(2!-S_1=2-1=1\).
\(S_2=1+4=5\), so \(3!-S_2=6-5=1\).
\(S_3=1+4+18=23\), so \(4!-S_3=24-23=1\).
\(S_4=1+4+18+96=119\), so \(5!-S_4=120-119=1\).
This suggests the conjecture
\(S_n=(n+1)!-1\).
Now prove this by induction.
Let \(H_n\) be the statement \(S_n=(n+1)!-1\).
Base case: when \(n=1\),
\(S_1=1=2!-1\),
so \(H_1\) is true.
Inductive step: assume \(H_k\) is true for some positive integer \(k\), so
\(S_k=(k+1)!-1\).
Then
\(S_{k+1}=S_k+u_{k+1}\)
\(=(k+1)!-1+(k+1)(k+1)!\)
\(=(k+1)!\bigl(1+k+1\bigr)-1\)
\(=(k+1)!(k+2)-1\)
\(=(k+2)!-1\).
So \(H_k\Rightarrow H_{k+1}\).
Therefore, by mathematical induction, \(S_n=(n+1)!-1\) for all positive integers \(n\).
