Given $(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$ Find $\frac{a_7}{a_{13}}$ Given $$(2x^2+3x+4)^{10}=\sum_{r=0}^{20}a_rx^r$$ Find Value of $\frac{a_7}{a_{13}}$
My try:
I assumed $A=2x^2$,$B=3x$ and $C=4$
Then we have the following cases to collect coefficient of $x^7$:
Case $1.$ $A^3 \times B^1 \times C^6$
Case $2.$ $A^2 \times B^3 \times C^5$
Case $3.$ $A^1 \times B^5 \times C^4$
case $4.$ $A^0 \times B^7 \times C^3$
Using multinomial theorem we get Coefficient of $x^7$ as:
$$a_7=10! \times \left(\frac{2^33^14^6}{3!6!}+\frac{2^13^54^4}{1!5!4!}+\frac{2^23^34^5}{2!3!5!}+\frac{2^03^74^3}{3!7!}\right)$$
Like wise we need to find $a_{13}$
But is there any better way?

Given $$(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}x^k$$
Now using  $\displaystyle x\rightarrow \frac{2}{x}$
$$\Rightarrow 2^{10}(2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k}x^{20-k}$$
$$\Rightarrow (2x^2+3x+4)^{10}=\sum^{20}_{k=0}a_{k}2^{k-10}x^{20-k}$$
Now Comparing Coefficient of $x^{7}$ on both side, we get 
$$a_{7}=a_{13}\cdot 2^{3}\Rightarrow \frac{a_{7}}{a_{13}}=8.$$
