Question 1116770
Expanding the binomial power {{{"["}}}{{{4+(-x)}}}{{{"]"^(4n)}}}
we get {{{4n+1}}} terms.
 
Each term is a product of
a combinatorial number,
a power of {{{4}}} ,
and a power of {{{(-x)}}} ,
where the sum of exponents adds up to {{{4n}}},
and the combinatorial number is combinations of {{{4n}}}
taking as many at a time as one of the exponents.
 
For the term in {{{x^(4n)}}}, we get {{{4^0*(-x)^(4n)=x^(4n)}}} , multiplied by
combinations of {{{4n}}} taking {{{0}}} or {{{4n}}} at a time, which is {{{1}}} either way.
 
For the term in {{{x^(n)}}}, we get {{{4^(3n)*(-x)^(n)=4^(3n)*(-x)^n=(-1)^n*4^(3n)*x^n}}} , multiplied by
combinations of {{{4n}}} taking {{{n}}} (or {{{3n}}} ) at a time.
You may represent combinations of {{{4n}}} taking {{{n}}} (or {{{3n}}} ) at a time by symbols such as
{{{matrix(3,2," ",4n,C," "," " ,n)}}} , or as {{{(matrix(2,1,4n,n))}}} ,
, but that combinatorial number taking {{{n}}} or {{{3n}}} ,
is {{{(4n)!/((3n)!n!)}}}{{{"="}}}{{{4n*(4n-1)*"..."*(3n+1)/(n*(n-1)*n-2)*",,,"3*2*1)}}} either way.
So, the coefficient of {{{x^n}}} is
{{{(-1)^n*4^(3n)*(4n)!/((3n)!n!)}}}{{{"="}}}{{{(-1)^n*4^(3n)*4n*(4n-1)*"..."*(3n+1)/(n*(n-1)*n-2)*",,,"3*2*1)}}}