Not ordinary sum-1

Relevant wiki: Taylor Series - Problem Solving

Write the sum as:

$\displaystyle\large \sum _{ n=1 }^{ \infty }{ \dfrac { (n(n-1)(n-2))+(3n(n-1))+n}{ { 3 }^{ n }\cdot n! } }$

$=\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n(n-1)(n-2)}{ { 3 }^{ n }\cdot n! } }+3\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n(n-1)}{ { 3 }^{ n }\cdot n! } }+\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n}{ { 3 }^{ n }\cdot n! } }$

$=\displaystyle\sum _{ n=3 }^{ \infty }{ \dfrac { 1}{ { 3 }^{ n }\cdot ( n-3)! } }+3\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { 1}{ { 3 }^{ n }\cdot (n-2)! } }+\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { 1}{ { 3 }^{ n }\cdot (n-1)! } }$

$=\dfrac 1{27}\displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { \left(\frac 13\right)^{n-3}}{ ( n-3)! } }+\dfrac 1{3}\displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { \left(\frac 13\right)^{n-2}}{ ( n-2)! } }+\dfrac 13\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { \left(\frac 13\right)^{n-1}}{ (n-1)! } }$

$=\dfrac 1{27}e^{1/3}+\dfrac 13e^{1/3}+\dfrac 13e^{1/3}~~\left(\small{\because \color{#3D99F6}{e^x=\displaystyle\sum_{r=0}^{\infty}\dfrac{x^r}{r!}}}\right)$

$\large =\dfrac{19e^{1/3}\pi^0}{27}$

$\Large \therefore 3+27+19+1=\boxed{\color{#D61F06}{50}}$

Proceeding differently, observe that the sum equals $(e^{(e^x)/3})'''$ at $x=0$ .