The value of (2)/(1 !) + (2 + 4)/(2 !) + (2 + 4 + 6)/(3 !) + . is

Here,

${T}_{n} = \dfrac{n ( n + 1 )}{n !} = \dfrac{n - 1 + 2}{( n - 1 ) !} = \dfrac{1}{( n - 2 ) !} + \dfrac{2}{( n - 1 ) !}$

$∴ S = \mathop{∑}_{n = 1}^{\infty}{T}_{n} = \mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( n - 2 ) !} + 2\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( n - 1 ) !}$

$= e + 2e = 3e$