Let $(2x^2 +3x+4)^{10} = \sum _{r=0}^{20} a_rx^r$. Then find $\frac{a_7}{a_{13}}$ I tried using the multinomial theorem where the n’th term is
$$\frac{10!}{r_1!r_2!r_3!}\times 2^{r_1}3^{r_2} 4^{r_3} x^{2r_1 + r_2}$$
($r$ is an integer)
Where $$r_1+r_2+r_3=10$$ and
For first part ie. $a_7$
$$2r_1+r_2 =7$$
But I don’t have enough equations to solve for all the $r$. Where am I going wrong?
 A: Method without finding coefficient
given$$\color{blue}{(2x^2 +3x+4)^{10}} = \sum _{r=0}^{20} a_rx^r$$ replace $x\to \frac{2}{x}$ $$2^{10} \frac{{(2x^2+3x+4)}^{10}}{x^{20}}=\sum_{r=0}^{20} \frac{2^ra_r}{x^r}$$ $$\iff 2^{10}\color{blue}{{(2x^2+3x+4)}^{10}}=\sum_{r=0}^{20} 2^ra_r x^{20-r}$$ Now as this is an identity the coefficient of $x^7$ in the LHS should be same as in RHS $$2^{10}a_7=2^{13}a_{13}\implies \frac{a_7}{a_{13}}=?$$
A: You are not going wrong. There are different ways to create $x^{7}$ term. For instance, you may choose $x^2$ once and choose $x$ five times (look at Case(II) below).
So, the goal is to find all possible ways to satisfy your equation. In other words:
$$Case(I): {r_{1} = 0}, \space {r_{2}=7} $$
$$Case(II): {r_{1} = 1}, \space {r_{2}=5} $$
$$Case(III): {r_{1} = 2}, \space {r_{2}=3} $$
$$Case(IV):  {r_{1} = 3}, \space {r_{2}=1} $$
Now, for each case, let's calculate the coefficient behind $x^7$:
$$Case(I): {10 \choose 0}{10 \choose 7}{3 \choose 3} \times {2^{0}} \times {3^{7}} \times {4^{3}} $$
Do the same things for other cases and then sum them together.
