Let S_n = 1 . (n $-$ 1) + 2 . (n $-$ 2) + 3 . (n $-$ 3) + ..... + (n $-$ 1) . 1, n $\ge$ 4.

The sum $$\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$ is equal to :

Let T_r = r(n $-$ r)<br><br>T_r = nr $-$ r²<br><br>$$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{T_r} = \sum\limits_{r = 1}^n {(nr - {r^2})} } $$<br><br>${S_n} = {{n\,.\,(n)(n + 1)} \over 2} - {{n(n + 1)(2n + 1)} \over 6}$<br><br>${S_n} = {{n(n - 1)(n + 1)} \over 6}$<br><br>Now, $$\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$<br><br>$$ = \sum\limits_{r = 4}^\infty {\left( {2.{{n(n - 1)(n + 1)} \over {6\,.\,n(n - 1)(n - 2)!}} - {1 \over {(n - 2)!}}} \right)} $$<br><br>$$ = \sum\limits_{r = 4}^\infty {\left( {{1 \over 3}\left( {{{n - 2 + 3} \over {(n - 2)!}}} \right) - {1 \over {(n - 2)!}}} \right)} $$<br><br>$$ = \sum\limits_{r = 4}^\infty {{1 \over 3}.{1 \over {(n - 3)!}} = {1 \over 3}(e - 1)} $$<br><br>Option (a)