A) \[\frac{S+1}{n!}\in \]integer
B) \[\frac{S+1}{n!}\notin \] Integer
C) \[\frac{S+1}{n!}\] cannot be discussed
D) None of these
Correct Answer: A
Solution :
[a] We have, \[S=\sum\limits_{k=1}^{n}{k(k!)}=\sum\limits_{k=1}^{n}{\{(k+1)-1\}}(k!)\] \[=\sum\limits_{k=1}^{n}{\{(k+1)!-k!\}=(n+1)!-1\Rightarrow S+1=(n+1)!}\] Thus, \[\frac{S+1}{n!}\in \] integer.
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