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?
 A: This approach is from the idea of Pascal's triangle. We write
$$
(2x^2+3x+4)^n = \sum_r a_{n,r} x^r
$$
Then 
$$
(2x^2+3x+4)^{n+1}=(\sum_r a_{n,r} x^r )(2x^2+3x+4) = \sum_r (2a_{n,r-2}+3a_{n,r-1}+4a_{n,r})x^r.$$
Thus, by comparing coefficients, we obtain a recurrence relation
$$
a_{n+1,r}=2a_{n,r-2}+3a_{n,r-1}+4a_{n,r}
$$
with an initial condition $a_{0,0}=1$, $a_{1,2}=2$, $a_{1,1}=3$, and $a_{1,0}=4$. 
I am not sure if this is a better way. But, it will be more convenient than computing factorials many times. Especially for small numbers, this method will give answers quickly. 
However, for large numbers, the number of recursion steps will increase enormously with $n$. 
A: 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 
$$a_{7}=a_{13}\cdot 2^{3}\Rightarrow \frac{a_{7}}{a_{13}}=8.$$
