Let Sn = 1 .

Question

Let Sn = 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 :

Accepted Answer

Let Tr = r(n - r)

Tr = nr - r2

$$ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {{T_r} = \sum\limits_{r = 1}^n {(nr - {r^2})} } $$

$${S_n} = {{n\,.\,(n)(n + 1)} \over 2} - {{n(n + 1)(2n + 1)} \over 6}$$

$${S_n} = {{n(n - 1)(n + 1)} \over 6}$$

Now, $$\sum\limits_{n = 4}^\infty  {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $$

$$ = \sum\limits_{r = 4}^\infty  {\left( {2.{{n(n - 1)(n + 1)} \over {6\,.\,n(n - 1)(n - 2)!}} - {1 \over {(n - 2)!}}} \right)} $$

$$ = \sum\limits_{r = 4}^\infty  {\left( {{1 \over 3}\left( {{{n - 2 + 3} \over {(n - 2)!}}} \right) - {1 \over {(n - 2)!}}} \right)} $$

$$ = \sum\limits_{r = 4}^\infty  {{1 \over 3}.{1 \over {(n - 3)!}} = {1 \over 3}(e - 1)} $$

Option (a)

Solutions · 1