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Question:

In the Binomial expansion of $(a-\frac x2)^6$ the coefficient of $x^3$ is 120 times the coeffiecient of $x^5$. Find the possible values of constant $a$.

I've done so far:

Using the fact that:

$${n \choose r} a^{n-r}\left(\frac{-x}{2}\right)^r$$ Equals to the term where $r$ is the coefficient of $x$ and $n$ is the degree of the binomial

I've reached the following equation: $$\left(\frac{-20a^3x^3}{960}\right)=\left(\frac{-6a^2x^5}{32}\right)$$

How can I use this to find a value of $a$?

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    $\begingroup$ That last equation is not correct; just remove the $x$'s to get the relation for the coefficients. $\endgroup$ – emacs drives me nuts Mar 8 '20 at 19:52
  • $\begingroup$ You must have $\dfrac{5a^3}{2}=\dfrac{120\cdot3a}{16}$ which gives $a=\pm3$. If I am not wrong....... $\endgroup$ – Piquito Mar 8 '20 at 19:54
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You have solved it correctly. Just compare the coefficients and not the terms including x. $$(\frac{-20a^3}{960})=(\frac{-6a}{32})$$ find a=$\pm$3.

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The coefficient of $x^3$ is $c_3:=\binom{6}{3}(\frac{-1}{2})^3 a^3= \frac{-20a^3}{8}$;

For $x^5$ this is $c_5:=\binom{6}{5}a (\frac{-1}{2})^5 = \frac{-6a}{32}$

Given is that $c_3 = 120c_5$ so

$$\frac{-20a^3}{8} = \frac{120\cdot -6a}{32}$$

Multiply both sides by $-32$ to get

$$80a^3 = 720a$$

divide by $80a$ on both sides to get

$$a^2 = 9$$

$a=\pm3$

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    $\begingroup$ You miss solution $a=0$. $\endgroup$ – emacs drives me nuts Mar 8 '20 at 21:19
  • $\begingroup$ @emacsdrivesmenuts yep, when I blithely divide by $80a$. $\endgroup$ – Henno Brandsma Mar 8 '20 at 21:29
  • $\begingroup$ Like the other 2 computations :-/ $\endgroup$ – emacs drives me nuts Mar 8 '20 at 21:49

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