# Binomial expansion of $(a-\frac x2)^6$

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 $$a$$?

• That last equation is not correct; just remove the $x$'s to get the relation for the coefficients. – emacs drives me nuts Mar 8 '20 at 19:52
• You must have $\dfrac{5a^3}{2}=\dfrac{120\cdot3a}{16}$ which gives $a=\pm3$. If I am not wrong....... – Piquito Mar 8 '20 at 19:54

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.

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

$$a^2 = 9$$

$$a=\pm3$$

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