Question

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

Answer 1

8

Answer 2

We need the coefficients $a_7$ and $a_{13}$ in

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

Write the expansion by choosing, for each term, nonnegative integers $i,j,k$ with

$$i+j+k=10,\quad \text{and total exponent }2i+j=r.$$

Thus:

$$a_r=\sum_{2i+j=r,\; i+j\le10}\frac{10!}{i!\,j!\,k!}\,2^i\,3^j\,4^k,\quad \text{with }k=10-i-j.$$

For $a_7$:

We require

$$2i+j=7,\quad j=7-2i,\quad \text{and}\quad k=10-(i+j)=10-(7-i)=3+i.$$

Valid choices (with $i$ such that $j\ge0$) are:

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

Thus,

$$a_7=\sum_{i=0}^{3}\frac{10!}{\,i!(7-2i)!(3+i)!}\,2^i\,3^{7-2i}\,4^{3+i}.$$

For $a_{13}$:

We require

$$2i+j=13,\quad j=13-2i,\quad \text{and}\quad k=10-(i+j)=10-(13-i)=i-3.$$

Since $k\ge0$ we need $i\ge 3$ and $j\ge0$ implies $i\le6$. Write $i= i'+3$ (with $i'=0,1,2,3$); then

$$j=13-2(i'+3)=7-2i',\quad k=(i'+3)-3=i'.$$

So,

$$a_{13}=\sum_{i'=0}^{3}\frac{10!}{\,(i'+3)!(7-2i')! \,i'!}\,2^{\,i'+3}\,3^{7-2i'}\,4^{\,i'}.$$

Observation: Re-indexing shows the same sum appears in both coefficients:

Let

$$S=\sum_{i=0}^{3}\frac{10!}{\,(i+3)!(7-2i)! \,i!}\,2^{i}\,3^{7-2i}\,4^{\,i}.$$

Then,

$$a_7=\underbrace{4^3}_{=64}\,\sum_{i=0}^{3}\frac{10!}{i!(7-2i)!(i+3)!}\,(2\cdot4)^i\,3^{7-2i} = 4^3\,S,$$ $$a_{13}=2^3\,S.$$

Thus,

$$\frac{a_7}{a_{13}}=\frac{4^3}{2^3}=\frac{64}{8}=8.$$

Summary:

• Core Explanation: Re-index the sum for $a_{13}$ with $i=i'+3$. Both coefficients share a common sum $S$ so that $a_7=4^3S$ and $a_{13}=2^3S$. Their ratio is $\frac{4^3}{2^3}=8$.