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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?

fitzmerl duron
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    $n=12$ and $C=1^6 2^6 \dbinom{12}{6}=59136$ – Raffaele Aug 31 '17 at 15:50
  • @Raffaele I wonder how that equation would fare well if one of them ($u$'s and $v$'s) or both had negative exponents. Is it applicable too for negative exponents? or maybe modify your equation a bit... – fitzmerl duron Aug 31 '17 at 15:54
  • No! It works just in this case because I was lucky to see that power $12$ did the job. It's not a proof so is not an answer, actually :) – Raffaele Aug 31 '17 at 16:00
  • @Raffaele It might be a good shortcut if $u$'s and $v$'s exponents were positive integers.... – fitzmerl duron Aug 31 '17 at 16:04

2 Answers2

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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, $C=\binom{12}{6}2^{12-6}=59136$

user362325
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If the middle term of the expansion of $(x^3 + 2y^2)^n$ is $C x^{18} y^{12},$ find C.

You want the coefficient of the term $\binom{6+6}{6}(x^3)^{6}(2y^2)^6$, so ....$$C~=~ 2^6 \binom{6+6}{6} ~=~59\,136$$

Therefore: okay $\checkmark$.

Graham Kemp
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