Question
If $$S = \left( 1 \right)\left( {1!} \right) + \left( 2 \right)\left( {2!} \right) + \left( 3 \right)\left( {3!} \right) + ..... + n\left( {n!} \right),$$ then
A.
$$\frac{{S + 1}}{{n!}} \in {\text{integer}}$$
B.
$$\frac{{S + 1}}{{n!}} \notin {\text{integer}}$$
C.
$$\frac{{S + 1}}{{n!}}{\text{ cannot be discussed}}$$
D.
None of these
Answer :
$$\frac{{S + 1}}{{n!}} \in {\text{integer}}$$
Solution :
We have,
$$\eqalign{
& S = \sum\limits_{k = 1}^n {k\left( {k!} \right)} = \sum\limits_{k = 1}^n {\left\{ {\left( {k + 1} \right) - 1} \right\}\left( {k!} \right)} \cr
& = \sum\limits_{k = 1}^n {\left\{ {\left( {k + 1} \right)! - k!} \right\} = \left( {n + 1} \right)! - 1} \cr
& \Rightarrow S + 1 = \left( {n + 1} \right)! \cr
& {\text{Thus, }}\frac{{S + 1}}{{n!}} \in {\text{integer}}{\text{.}} \cr} $$