Finding certain coefficients in Taylor expansion of $ \log(1 +qx^2 + rx^3)$ This exercise is part of the STEP $3$ paper from $2014$. 
At a certain point in the problem, we 're supposed find $a_n$ for $n = {2,5,7,9}$
where $a_n$ is the coefficient of $x^n$ in the series expansion around $0$ of $ \log(1+qx^2 + rx^3)$.
I knew at first sight that this can't be as direct as it seems, precisely because it'd be tedious to expand all the powers of $(qx^2+rx^3)$ to find the coefficients we're looking for. When I saw I couldn't get anywhere with this,
I looked at the provided 'hints and solutions' and , to my surprise, this is all they said, 'the coefficients may be found easily by expanding $ \log(1 +qx^2 + rx^3)$' . '
While using a computational resource such as Wolfram | Alpha may find the solution quickly, how does one do it by hand?

I think the suggested solution is straightforward enough not to be dismissed out-of-hand.
We know
$$
\log(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} \pm\dotsm
$$
Let $u = qx^2 + rx^3$.  Then
$$
\log(1+qx^2 + rx^3) = (qx^2 + rx^3) - \frac{1}{2}(qx^2 + rx^3)^2 + \frac{1}{3}(qx^2 + rx^3)^3 + \dotsm
$$
Only the first power of $(qx^2 + rx^3)$ is going to produce a term of degree $2$.  So $a_2 = q$.  The expansion of $(qx^2 + rx^3)^2$ is going to produce terms of degree $4$, $5$, and $6$, and the expansion of $(qx^2 + rx^3)^3$ is going to produce terms of degree $6$, $7$, $8$, and $9$.  The expansion of $(qx^2+rx^3)^4$ is going to produce terms of degree $8$, $9$, and higher.  The expansion of $(qx^2 + rx^3)^5$ is going to produce terms of degree $10$ and higher.
So $a_5$ comes from the cross term in the second binomial power:
$$
    a_5 = -\frac{1}{2} (2qr) = -qr
$$
The degree $7$ term comes from  $(qx^2)^2(rx^3)^1$.  From the binomial theorem there is a factor of $3$, but there is also the $\frac{1}{3}$ in front of the power to cancel those.  Therefore
$$
    a_7 = q^2 r
$$
The degree $9$ term comes from $(qx^2)^0(rx^3)^3$ in the third power and $(qx^2)^3(rx^3)^1$ in the fourth power.  So
$$
    a_9 = \frac{1}{3}r^3 - \frac{1}{4}\cdot 4 q^3r = \frac{1}{3}r^3 - q^3 r
$$
