Let ${\left(2x^2+3x+4\right)}^{10}=\sum _{r=0}^{20} a_r{x}^r\text{.}$Then ,  $\frac{a_7}{a_{13}}$is equal 

The general term in the expansion of ${\left(2x^2+3x+4\right)}^{10}$, is .

$\frac{10!}{r!s!t!}{\left(2x^2\right)}^r(3x{)}^s{4}^t=\frac{10!}{r!s!t!}{2}^r\cdot 3^s\cdot 4^t{x}^{2r+s}$ 

where $r+s+t=10$ 

Clearly , $a_7=$Coefficient of $x^7$and $a_{13}=$Coefficient of $x^{13}$

For the coefficient of $x^7$ , we must have 

$2r+s=7$and $r+s+t=10$ 

$\Rightarrow r=0,s=7,t=3;r=1,s=5,t=4;r=2,s=3,t=5$ and $r=3,s=1,t=6$ 

$\therefore a_7=\frac{10!}{1!7!3!}{2}^0\cdot 3^7\cdot 4^3+\frac{10!}{1!5!4!}2\cdot 3^5\cdot 4^4$ $+\frac{10!}{2!3!5!}{2}^2\cdot 3^3\cdot 4^5+\frac{10!}{3!1!6!}{2}^3\cdot 3^1\cdot 4^6$

$\Rightarrow a_7=\frac{10!\cdot 3^7\cdot 4^3}{7!3!}+\frac{10!\cdot 2\cdot 3^5\cdot 4^4}{5!4!}+\frac{10!\cdot 2^2\cdot 3^3\cdot 4^5}{2!3!5!}+\frac{10!\cdot 2^3\cdot 3\cdot 4^6}{3!6!}$

For the coefficient of $x^{13}$, we must have  

$2r+s=13$ and $r+s+t=10$ 

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}r=3,s=7,t=0;r=4,s=5,t=1;r=5,s=3,t=2;\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}r=6,s=1,t=3\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}_{13}=\frac{10!\cdot 2^3\cdot 3^7\cdot 4^0}{3!7!0!}+\frac{10!\cdot 2^4\cdot 3^5\cdot 4^1}{4!5!1!}+\frac{10!\cdot 2^5\cdot 3^3\cdot 4^2}{5!3!2!}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}_{13}=\frac{10!\cdot 2^6\cdot 3^1{4}^3}{6!1!3!}+\frac{10!\cdot 2^6\cdot 3\cdot 4^3}{6!3!}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

$\Rightarrow a_{13}=\frac{10!\cdot 2^3\cdot 3^7}{3!7!}+\frac{10!\cdot 3^5\cdot 2^6}{4!5!}+\frac{10!\cdot 3^3\cdot 2^9}{5!3!2!}+\frac{10!\cdot 3\cdot 2^{12}}{6!3!}$

Clearly , $a_7=2^3{a}_{13}$ 

$\Rightarrow \frac{a_7}{a_{13}}=8$