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.

Binomial expansion \begin{eqnarray} (2x-4y)^7= \cdots +\binom{7}{3}(2x)^3 (-4y)^4 +\cdots \end{eqnarray} — Donald Splutterwit, Sep 11 '18 at 01:26
How did you choose that combination and what would the final answer be? — Murph Jones, Sep 11 '18 at 01:29
\begin{eqnarray} (a+b)^n= \cdots +\binom{n}{r}a^r b^{n-r} +\cdots \end{eqnarray} What term do you need to choose to obtain the monomial $x^3 y^4$ ? ... simplify ? You tell me what you think. $\ddot \smile$ — Donald Splutterwit, Sep 11 '18 at 01:34
n=7 and r would have to equal 3. I just don't understand how this summation (if it is a summation) works — Murph Jones, Sep 11 '18 at 01:45
I do understand how you assigned values to r and n the question now is just how the recipe works — Murph Jones, Sep 11 '18 at 01:46
Try expanding $(a+b)^2$ and $(a+b)^3$ ... look carefully at the terms ... now have a look at https://en.wikipedia.org/wiki/Binomial_theorem — Donald Splutterwit, Sep 11 '18 at 01:51

score 0 · Answer 1 · answered Sep 11 '18 at 01:54

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.

score 0 · Answer 2 · answered Sep 18 '18 at 12:30

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 $x^4$.

Binomial Expansion coefficients

2 Answers2