1. We need the coefficients of $x^7$ and $x^{13}$ in

$$(2x^2+3x+4)^{10}.$$

Let

$$(2x^2+3x+4)^{10}= \sum_{r=0}^{20} a_r x^r.$$

We must find

$$\frac{a_7}{a_{13}}.$$

2. Use symmetry by rewriting the polynomial in reciprocal form.

Observe:

$$2x^2+3x+4=x^2\left(2+\frac{3}{x}+\frac{4}{x^2}\right).$$

So

$$(2x^2+3x+4)^{10}=x^{20}\left(2+\frac{3}{x}+\frac{4}{x^2}\right)^{10}.$$

Now factor inside as

$$2+\frac{3}{x}+\frac{4}{x^2}=\frac{1}{x^2}(2x^2+3x+4).$$

But a more useful identity is:

$$2x^2+3x+4 = x^2\left(4\left(\frac1x\right)^2+3\left(\frac1x\right)+2\right).$$

Hence

$$(2x^2+3x+4)^{10}=x^{20}(4x^{-2}+3x^{-1}+2)^{10}.$$

Therefore, coefficient comparison gives a symmetry: coefficient of $x^r$ in $(2x^2+3x+4)^{10}$ equals coefficient of $x^{20-r}$ in $(4x^2+3x+2)^{10}$.

3. Now use the general coefficient formula.

To get $x^r$ in

$$(2x^2+3x+4)^{10},$$

choose from the 10 factors:

• $i$ times the term $2x^2$,
• $j$ times the term $3x$,
• $k$ times the term $4$, with

$$i+j+k=10$$

and power condition

$$2i+j=r.$$

Then

$$a_r=\sum \frac{10!}{i!j!k!}2^i3^j4^k,$$

where $k=10-i-j$ and $2i+j=r$.

4. Compute $a_7$.

We need

$$2i+j=7, \quad i+j+k=10.$$

Possible nonnegative integer solutions:

• $i=0, j=7, k=3$
• $i=1, j=5, k=4$
• $i=2, j=3, k=5$
• $i=3, j=1, k=6$

So

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

Now simplify each term:

• First term:

$$\frac{10!}{7!3!}3^7 4^3=120\cdot 2187\cdot 64=16796160.$$

• Second term:

$$\frac{10!}{1!5!4!}2\cdot 3^5\cdot 4^4=1260\cdot 2\cdot 243\cdot 256=156764160.$$

• Third term:

$$\frac{10!}{2!3!5!}2^2\cdot 3^3\cdot 4^5=2520\cdot 4\cdot 27\cdot 1024=278691840.$$

• Fourth term:

$$\frac{10!}{3!1!6!}2^3\cdot 3\cdot 4^6=840\cdot 8\cdot 3\cdot 4096=82575360.$$

Thus

$$a_7=16796160+156764160+278691840+82575360=534827520.$$

5. Compute $a_{13}$.

We need

$$2i+j=13, \quad i+j+k=10.$$

Possible solutions:

• $i=3, j=7, k=0$
• $i=4, j=5, k=1$
• $i=5, j=3, k=2$
• $i=6, j=1, k=3$

So

$$a_{13}=\frac{10!}{3!7!0!}2^3 3^7 4^0 +\frac{10!}{4!5!1!}2^4 3^5 4^1 +\frac{10!}{5!3!2!}2^5 3^3 4^2 +\frac{10!}{6!1!3!}2^6 3^1 4^3.$$

Now simplify:

• First term:

$$\frac{10!}{3!7!}2^3 3^7=120\cdot 8\cdot 2187=2099520.$$

• Second term:

$$\frac{10!}{4!5!1!}2^4 3^5 4=1260\cdot 16\cdot 243\cdot 4=19595520.$$

• Third term:

$$\frac{10!}{5!3!2!}2^5 3^3 4^2=2520\cdot 32\cdot 27\cdot 16=34836480.$$

• Fourth term:

$$\frac{10!}{6!1!3!}2^6 3\cdot 4^3=840\cdot 64\cdot 3\cdot 64=10321920.$$

Thus

$$a_{13}=2099520+19595520+34836480+10321920=66853440.$$

6. Now take the ratio:

$$\frac{a_7}{a_{13}}=\frac{534827520}{66853440}=8.$$

7. Final answer:

$$\boxed{8}$$