Not ordinary sum-2

Relevant wiki: Taylor Series Manipulation

Write the sum as:

$\displaystyle\large \sum _{ n=1 }^{ \infty }{ \dfrac { (n(n-1)(n-2)(n-3)+(6n(n-1)(n-2))+(7n(n-1))+n}{ { 4 }^{ n }\cdot n! } }$

$=\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n(n-1)(n-2)(n-3)}{ { 4 }^{ n }\cdot n! } }+6\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n(n-1)(n-2)}{ { 4 }^{ n }\cdot n! } }+7\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n(n-1)}{ { 4}^{ n }\cdot n! } }+\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { n}{ {4}^{ n }\cdot n! } }$

$=\displaystyle\sum _{ n=4 }^{ \infty }{ \dfrac {1}{ { 4 }^{ n }\cdot (n-4)! } }+6\displaystyle\sum _{ n=3 }^{ \infty }{ \dfrac { 1}{ { 4 }^{ n }\cdot ( n-3)! } }+7\displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { 1}{ { 4 }^{ n }\cdot (n-2)! } }+\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { 1}{ { 4 }^{ n }\cdot (n-1)! } }$

$=\dfrac{1}{4^4}\displaystyle\sum _{ n=4 }^{ \infty }{ \dfrac { \left(\frac 14\right)^{n-4}}{ (n-4)! } }+\dfrac 6{4^3}\displaystyle\sum _{ n=3 }^{ \infty }{ \dfrac { \left(\frac 14\right)^{n-3}}{ ( n-3)! } }+\dfrac 7{4^2}\displaystyle\sum _{ n=2 }^{ \infty }{ \dfrac { \left(\frac 14\right)^{n-2}}{ ( n-2)! } }+\dfrac 14\displaystyle\sum _{ n=1 }^{ \infty }{ \dfrac { \left(\frac 14\right)^{n-1}}{ (n-1)! } }$

$=\dfrac 1{4^4}e^{1/4}+\dfrac 6{4^3}e^{1/4}+\dfrac 7{4^2}e^{1/4}+\dfrac 14e^{1/4}~~\left(\small{\because \color{#3D99F6}{e^x=\displaystyle\sum_{r=0}^{\infty}\dfrac{x^r}{r!}}}\right)$

$\large =\dfrac{201e^{1/4}\pi^0}{256}$

$\Large \therefore 4+256+201+1=\boxed{\color{#D61F06}{462}}$