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.

  • $\begingroup$ Without multinomial theorem, this would be tedious at best, case work at worst. $\endgroup$ – nathan.j.mcdougall Apr 17 '15 at 4:33
  • $\begingroup$ How would you approach this with the multinomial theorem then? $\endgroup$ – rawrrawr Apr 17 '15 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 $40095+792=40887$


$$ \sum_{n=0}^{12} \ _{12}C_n 1^{12-n} (3x+2x^3)^n = \sum_{n=0}^{12} \ _{12}C_n \sum_{k=0}^n\ _nC_k (3x)^k(2x^3)^{n-k} $$

so that $$ n-k =0,\ k=4,\ {\rm or}\ n-k=1,\ k=3 $$

Hence $$ _{12}C_4\ _4C_4 3^4 + \ _{12}C_4\ _4C_3 3^32 $$


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