# Finding the coefficient of $x^2$ in an expansion

I was given this question as a practise assignment and I am unsure of my answer.

The coefficient of $$x^2$$ in the expansion of $$(x+\frac{1}{ax})^8$$ is 7. Find the possible value of $$a$$.

I did $$(x+\frac{1}{ax})^8$$ = $$(x (1+\frac{1}{ax^2}))^8$$

Given my answer, does that mean that $$a=7$$?

Edit: Thank you for everyone's help! I think I understand it a little bit now. I will have to review this question again and practise more.

• You are expected to use the binomial expansion. Mar 16, 2021 at 22:30
• No. What is the expression for the coefficient of $x^2$ in the expansion? Mar 16, 2021 at 22:38
• If you want to split the problem like that $(x+\frac{1}{ax})^8 = (x(1+\frac{1}{ax^2}))^8$, you have to remember that you're looking for the coefficient of $x^2$, $x^8 \cdot (1+\frac{1}{a^2})^8$ here you'll use coefficient of $\frac{1}{x^6}$, which is just as difficult as expanding it all, so use binomial Mar 16, 2021 at 23:15

hint

$$(x+\frac{1}{ax})^2=x^2+\frac{1}{a^2x^2}+\frac{2}{a}$$

So, you just need to find the coefficient of $$X$$ in the expansion

$$(X+\frac{1}{a^2X}+\frac{2}{a})^4$$

• So if $a^2 =7$, $a=\sqrt{7}$? Mar 16, 2021 at 22:37

Two options:

1. With the binomial theorem: the expansion of $$(a+b)^n$$ is $$(a+b)^n = \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2+\cdots + \binom{n}{n-1}a^1b^{n-1} + \binom{n}{n}a^0b^n.$$ Set $$a=x$$, $$b=\frac{1}{ax}$$, $$n=8$$, and figure out which of the terms is the $$x^2$$ term. This will give you an expression involving $$a$$, which you can then solve for $$a$$.

2. With derivatives. Rewrite the binomial as $$\left(x + \frac{1}{ax}\right) = \frac{1}{x}\left(x^2 + \frac{1}{a}\right).$$ Therefore, $$\left(x + \frac{1}{ax}\right)^8 = \frac{1}{x^8}\left(x^2+\frac{1}{a}\right)^8.$$ The coefficient of $$x^2$$ will be the coefficient of $$x^{10}$$ in $$(x^2+\frac{1}{a})^{8}$$.

This can be found using derivatives: if $$p(x)$$ is a polynomial, then the constant term of $$p^{(k)}(x)$$ is $$k!$$ times the coefficient of $$x^k$$ in $$p(x)$$. So $$p^{(k)}(0)$$ will give you $$k!$$ times the coefficient of $$x^k$$ in $$p(x)$$. Replace $$x^2$$ with $$y$$, and look for the coefficient of $$y^5$$ in the expansion of $$(y+\frac{1}{a})^8$$ to find the coefficient of $$x^{10}$$ in the original, and from there you can get the one for $$x^2$$ in the original expression. This will give you an expression involving $$a$$, which you can then solve for $$a$$.

• Using the formula in option 2, the coefficient of $x^{10}$ is $\frac{56}{a^3}$ so $\frac{56}{a^3} = 7$ and $a=2$? Mar 16, 2021 at 23:15
• @MissRose: It’s not my job to grade your homework; I told you two different ways to approach it, it is now up to you. Mar 17, 2021 at 0:03

Write the expression as $${1 \over a^8 x^8} (1+a x^2)^8$$ and find the coefficient of $$x^{10}$$ in $$(1+a x^2)^8$$.

Use the binomial theorem to compute the coefficient of $$x^{10}$$ in $$(1+a x^2)^8$$. Hint: It is the 5th coefficient.

Call this expression $$E$$ (it will be a formula involving $$a$$ and some numbers). Then the coefficient of $$x^{2}$$ in the original expression is $${1 \over a^8}E$$.

Then solve the expression $${1 \over a^8}E = 7$$ for $$a$$. (You will need to replace $$E$$ by the expression you computed first before solving.)

• I have no idea where you are getting that from. Mar 16, 2021 at 22:45
• Using your formula $(1+ax^2)^8$, the coeffient of the $x^{10}$ is $56a^5x^{10}$ so $56a^5=7$ and $a=\sqrt[5]{\frac {1}{8}}$? Mar 16, 2021 at 23:09
• Close, you have to find the coefficient of $x$ in the original formula. So the expression is ${56 \over a^3} = 7$. Mar 16, 2021 at 23:14

You want to solve

$${8 \choose k}x^{8-k}\left(\frac{1}{ax^2}\right)^k=7x^2$$

This can be expressed in the form

$${8 \choose k}\cdot\frac{1}{a^k}x^{8-3k}=7x^2$$

Solving $$8-3k=2$$ gives $$k=2$$. Solving $$\displaystyle{8\choose2}\cdot\dfrac{1}{a^2}=7x^2$$ gives $$a^2=4$$.

However, $$-2$$ turns out to be extraneous, so $$2$$ is the only solution.

• @copper.hat You are correct, I will amend my answer. Mar 16, 2021 at 23:19