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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.

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  • $\begingroup$ You are expected to use the binomial expansion. $\endgroup$ Mar 16, 2021 at 22:30
  • $\begingroup$ No. What is the expression for the coefficient of $x^2$ in the expansion? $\endgroup$
    – copper.hat
    Mar 16, 2021 at 22:38
  • $\begingroup$ 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 $\endgroup$ Mar 16, 2021 at 23:15

4 Answers 4

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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

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

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  • $\begingroup$ So if $a^2 =7$, $a=\sqrt{7}$? $\endgroup$
    – Miss Rose
    Mar 16, 2021 at 22:37
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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 $a$.

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  • $\begingroup$ Using the formula in option 2, the coefficient of $x^{10}$ is $\frac{56}{a^3}$ so $\frac{56}{a^3} = 7$ and $a=2$? $\endgroup$
    – Miss Rose
    Mar 16, 2021 at 23:15
  • $\begingroup$ @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. $\endgroup$ Mar 17, 2021 at 0:03
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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.)

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

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

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