<s>[INST] Finding the Coefficient of $(x^3 + 2y^2)^n$ containing $x^{18} y^{12}$ I was helping somebody when I scratched my head because of this question. It goes like 
this:

If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.

My work:
I let $u = x^3$ and $v = 2y^2.$ Then doing this: $$(u)^6 = (x^3)^6 \space  and \space \space \left(\frac{v}{2} \right)^6 = (y^2)^6$$
we get $u^6 = x^{18}$ and $\left(\frac{v^6}{64} \right) = y^{12}$
We now conclude that the expression $(u + v)^n$ has a term $(u^6)\left(\frac{v^6}{64} \right)$ along its expansion when $u = x^3$ and $v = 2y^2$ We need to find its equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ when we go back to dealing with $(x^3 + 2y^2)^n.$
Everybody knows that in the binomial expansion of $(u + v)^n,$ in each term, the sum of the exponents of $u$ and $v$ is $n.$ and there are $n+1$ terms. 
With that in mind, the sum of the particular term $(u^6)\left(\frac{v^6}{64} \right)$ is $n = 12$ and the number of terms in that particular expansion is $12+1 = 13.$ Since the problem asks for the coefficient of the middle term $C x^{18} y^{12},$ we need to find its middle term. Turns out, in the binomial expansion containing $13$ terms, the middle term would 
be the $7$th term.
Now looking for for the expression of the $7$th term:
$$nth \space term = C(n, r-1) u^{n-r+1} v^{r-1}$$
$$expression \space of \space 7th \space term = C(12, 7-1) (x^3)^{12-7+1} (2y^2)^{7-1}$$
$$ = (924) (x^3)^{6} (2y^2)^{6}$$
$$ = (924) (x^{18}) (2)^{6}(y^2)^{6}$$
$$ = (924) (x^{18}) (64) (y^2)^{6}$$
$$ = (924) (x^{18}) (64) (y^{12})$$
$$ = 59136 x^{18} y^{12}$$
Therefore, we conclude that $C = 59136.$
Lastly, the equivalent term of $(u^6)\left(\frac{v^6}{64} \right)$ from $(u + v)^n$, when we go back to dealing with $(x^3 + 2y^2)^n$, is $59136x^{18}y^{12}$
I've done my best but I couldn't verify it. Is my solution correct?
 [/INST] The general term of the expansion $(x^3+2y^2)^n$ is
$$\binom{n}{r}2^{n-r}x^{3r}y^{2n-2r}$$
For the term of $x^{18}y^{12}$, take $r=6$ and such that $n=12$
So the coefficient, $C=\binom{12}{6}2^{12-6}=59136$
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