Find the coefficient of $x^4$ in the expansion of $(1 + 3x + 2x^3)^{12}$?

I have not learnt the multinomial theorem yet, and was trying to approach this using the binomial theorem. I divided the terms as $a$ being $(1+3x)$ and $b$ being $2x^3$.

I then used $${12\choose 3}(1+3x)^8(2x^3)^3$$. However now do I have to expand the second term again?

Please let me know where I am getting wrong. It seems there should be an easier way to approach this problem.

• Without multinomial theorem, this would be tedious at best, case work at worst. Apr 17, 2015 at 4:33
• How would you approach this with the multinomial theorem then? Apr 17, 2015 at 4:34

With multinomial theorem, we want the coefficient on $x^4$ terms. There are two terms that contain $x^4$. These are:
• $1^{10}\cdot(3x)^1\cdot (2x^3)^1$
• $1^{8}\cdot(3x)^4\cdot (2x^3)^0$
With coefficients respectively of: $$3\cdot 2\cdot \binom {12} {10,1,1}=\frac{6\cdot12!}{10!\cdot 1!\cdot 1!}=792$$ $$3^4\cdot\binom {12} {8,4,0}=\frac{81\cdot12!}{8!\cdot 4!\cdot 0!}=40095$$ Which then add to give your coefficient of $x^4$ of $40095+792=40887$
$$\sum_{n=0}^{12} \binom{12}{n} 1^{12-n} (3x+2x^3)^n = \sum_{n=0}^{12} \binom{12}{n} \sum_{k=0}^n \binom{n}{k} (3x)^k(2x^3)^{n-k}$$
so that $$n-k =0,\ k=4,\ {\rm or}\ n-k=1,\ k=1$$
Hence $$\binom{12}{4}\binom{4}{4} 3^4 + \binom{12}{2}\binom{2}{1} (3^1)2$$