3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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 $40095+792=40887$

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.