4, 3, 2, 1: Part 2

Write the sum as:

$\displaystyle\large \dfrac 14\sum _{ n=1 }^{ \infty }{ \dfrac { ({ n }^{ 2 } -n)+n}{ { 3 }^{ n }\cdot n! } }$

$=\dfrac 14\left(\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n(n-1)}{ { 3 }^{ n }\cdot n! } }+\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n}{ { 3 }^{ n }\cdot n! } }\right)$

$=\dfrac 14\left(\displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { 1}{ { 3 }^{ n }\cdot ( n-2)! } }+\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { 1}{ { 3 }^{ n }\cdot (n-1)! } }\right)$

$=\dfrac 14\left(\dfrac 19\displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { \left(\frac 13\right)^{n-2}}{ ( n-2)! } }+\dfrac 13\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { \left(\frac 13\right)^{n-1}}{ (n-1)! } }\right)$

$=\dfrac 14\left(\dfrac 19e^{1/3}+\dfrac 13e^{1/3}\right)~~\left(\small{\because \color{#3D99F6}{e^x=\displaystyle\sum_{r=0}^{\infty}\dfrac{x^r}{r!}}}\right)$

$\large =\dfrac{e^{1/3}}{9}$

$\Large \therefore 9+3=\boxed{\color{#D61F06}{12}}$ $\huge\color{grey}{\star}$