Binomial Expansion coefficients

Question

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.

Answer 1

The Binomial Theorem states $$(a+b)^n= \sum_{r=0}^{n}\binom{n}{ r} (a)^{n-r}(b)^r$$ and so $$(2x-4y)^7=\sum_{r=0}^{7}\binom{7}{ r} (2x)^{7-r}(-4y)^r.$$

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

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 $$\binom{n}{r}(a)^{n-r}(b)^{r}$$ for that particular $r$, as I did above.

Answer 2

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

We obtain \begin{align*} \color{blue}{[x^3y^4]}&\color{blue}{(2x-4y)^7}\\ &=[x^3y^4]\sum_{k=0}^7\binom{7}{k}(2x)^k(-4y)^{7-k}\tag{1}\\ &=[y^4]\binom{7}{3}2^3(-4y)^4\tag{2}\\ &=\binom{7}{3}2^3(-4)^4\tag{3}\\ &\,\,\color{blue}{=71\,680} \end{align*}

Comment:

• In (1) we apply the binomial theorem.

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

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