Let $$(2x^2 + 3x + 4)^{10} = \sum_{r=0}^{20} a_r x^r$$. Then $$\frac{a_7}{a_{13}}$$ is equal to __________

We have to compare the coefficients of $$x^7$$ and $$x^{13}$$ in the expansion of $$(2x^2+3x+4)^{10}$$.

First recall the multinomial theorem, which states that for any positive integer $$n$$,

$$ (a+b+c)^n=\sum_{k_1+k_2+k_3=n}\dfrac{n!}{k_1!k_2!k_3!}\;a^{\,k_1}b^{\,k_2}c^{\,k_3}. $$

Here we identify

$$ a=2x^2,\qquad b=3x,\qquad c=4,\qquad n=10. $$

So the general term after expansion is

$$ \dfrac{10!}{k_1!k_2!k_3!}\,(2x^2)^{k_1}\,(3x)^{k_2}\,4^{\,k_3}, $$

where the exponents satisfy $$k_1+k_2+k_3=10$$ with each $$k_i\ge 0$$. This general term contributes

$$ \dfrac{10!}{k_1!k_2!k_3!}\,2^{k_1}3^{k_2}4^{k_3}\;x^{\,2k_1+k_2}. $$

Hence the power of $$x$$ in any term is determined by

$$ 2k_1+k_2=r. $$

We shall solve this relation twice, once for $$r=7$$ and once for $$r=13$$.

Case 1: $$r=7$$. Setting $$2k_1+k_2=7$$ with the additional condition $$k_1+k_2+k_3=10$$, we write $$k_1=t$$. Then

$$ k_2=7-2t,\qquad k_3=10-t-(7-2t)=3+t. $$

Non-negative integers give the admissible values

$$ t=0,1,2,3. $$

Consequently the four ordered triples are

$$ (k_1,k_2,k_3)=(0,7,3),(1,5,4),(2,3,5),(3,1,6). $$

Thus

$$ a_7=\sum_{t=0}^{3}\dfrac{10!}{k_1!k_2!k_3!}\;2^{k_1}3^{k_2}4^{k_3}. $$

Case 2: $$r=13$$. Now $$2k_1+k_2=13$$. Put again $$k_1=t$$; then

$$ k_2=13-2t,\qquad k_3=10-t-(13-2t)=t-3. $$

We require $$k_3\ge 0,$$ so $$t\ge 3$$. Allowable integer values are

$$ t=3,4,5,6. $$

This yields another set of four triples:

$$ (k_1,k_2,k_3)=(3,7,0),(4,5,1),(5,3,2),(6,1,3). $$

Therefore

$$ a_{13}=\sum_{t=3}^{6}\dfrac{10!}{k_1!k_2!k_3!}\;2^{k_1}3^{k_2}4^{k_3}. $$

Pairwise comparison of the two sets. Notice that each triple in the $$a_{13}$$ list is obtained from the corresponding triple in the $$a_7$$ list by the simple shift

$$ (k_1,k_2,k_3)\longrightarrow(k_1+3,\;k_2,\;k_3-3). $$

Let us compare one matched pair. Take a triple $$(k_1,k_2,k_3)$$ from the $$a_7$$ list and build the twin $$\bigl(k_1+3,k_2,k_3-3\bigr)$$ for $$a_{13}$$. The ratio of their individual contributions is

$$$ \frac{\displaystyle \dfrac{10!}{k_1!k_2!k_3!}\;2^{k_1}3^{k_2}4^{k_3}} {\displaystyle \dfrac{10!}{(k_1+3)!\,k_2!\,(k_3-3)!}\;2^{k_1+3}3^{k_2}4^{\,k_3-3}} = \frac{(k_1+3)!\,(k_3-3)!}{k_1!\,k_3!}\;\frac{2^{k_1}}{2^{k_1+3}}\;\frac{4^{k_3}}{4^{k_3-3}}. $$$

Now we simplify each factor:

• The factorial quotient is

$$\frac{(k_1+3)!\,(k_3-3)!}{k_1!\,k_3!}=\frac{(k_1+3)(k_1+2)(k_1+1)}{k_3(k_3-1)(k_3-2)}.$$

Because for every triple in the $$a_7$$ list we actually have $$k_3=k_1+3,$$ the numerator and the denominator above are identical, giving this whole quotient the value $$1$$.

• The powers of $$2$$ contribute $$2^{-3}=1/8.$$

• The powers of $$4$$ contribute $$4^{3}=64.$$

Multiplying these three simplified factors we obtain

$$ 1\times\frac{1}{8}\times 64=8. $$

This computation shows that each individual term present in $$a_7$$ is exactly eight times the corresponding term in $$a_{13}$$. Since both coefficients are sums of four such matched terms, the same factor $$8$$ carries over to the entire sums. Therefore

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

So, the answer is $$8$$.