Vtyjkyuk

Find the coefficient of $x^3 y^4$ in the expansion of $(2x-4y)^7$. I would also like an explanation for how the final answer was obtained.

The Binomial Theorem states
$$(a+b)^n= sum_r=0^nbinomn r (a)^n-r(b)^r$$
and so
$$(2x-4y)^7=sum_r=0^7binom7 r (2x)^7-r(-4y)^r.$$

So, we simply need to consider the $5$th term (i.e., when $r=4$):
$$binom7 4 (2x)^7-4(-4y)^4=binom742^34^4x^3y^4.$$

I’ll leave it to you to evaluate what this equals.

In general, whenever you are asked to find a particular coefficient in the Binomial expansion, try to reword the question into one which asks “What value of $r$ will give me the desired term?” and then evaluate $$binomnr(a)^n-r(b)^r$$ for that particular $r$, as I did above.

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.

We obtain
beginalign*
colorblue[x^3y^4]&colorblue(2x-4y)^7\
&=[x^3y^4]sum_k=0^7binom7k(2x)^k(-4y)^7-ktag1\
&=[y^4]binom732^3(-4y)^4tag2\
&=binom732^3(-4)^4tag3\
&,,colorblue=71,680
endalign*

Comment:

• In (1) we apply the binomial theorem.




• In (2) we select the coefficient of $x^3$.




• In (3) we select the coefficient of $x^4$.