SOLUTION: Find the Coefficient of X^n in the expansion of (4-x)^4n

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Question 1116770: Find the Coefficient of X^n in the expansion of (4-x)^4n
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Expanding the binomial power %22%5B%224%2B%28-x%29%22%5D%22%5E%284n%29
we get 4n%2B1 terms.

Each term is a product of
a combinatorial number,
a power of 4 ,
and a power of %28-x%29 ,
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%5E%284n%29, we get 4%5E0%2A%28-x%29%5E%284n%29=x%5E%284n%29 , multiplied by
combinations of 4n taking 0 or 4n at a time, which is 1 either way.

For the term in x%5E%28n%29, we get , 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%283%2C2%2C%22+%22%2C4n%2CC%2C%22+%22%2C%22+%22+%2Cn%29 , or as %28matrix%282%2C1%2C4n%2Cn%29%29 ,
, but that combinatorial number taking n or 3n ,
is %284n%29%21%2F%28%283n%29%21n%21%29%22=%22 either way.
So, the coefficient of x%5En is
%28-1%29%5En%2A4%5E%283n%29%2A%284n%29%21%2F%28%283n%29%21n%21%29%22=%22