MHB Finding the Coefficient Ratio of a Binomial Expansion Problem

[Saitama]

Problem:
Let $\displaystyle (3x^2+2x+c)^{12}=\sum_{r=0}^{24} A_rx^r$ and $\dfrac{A_{19}}{A_5}=\dfrac{1}{2^7}$, then $c$ is

A)4
B)5
C)6
D)7

Attempt:
One way I know of to solve such a problem is by using the multinomial theorem but I am reluctant to use it here. It would take a lot of time to write down all the combinations and perform the calculations if I use multinomial theorem. Since this is an exam problem, I feel there exists a much better method.

Any help is appreciated. Thanks!

 

[anemone]

Problem:
Let $\displaystyle (3x^2+2x+c)^{12}=\sum_{r=0}^{24} A_rx^r$ and $\dfrac{A_{19}}{A_5}=\dfrac{1}{2^7}$, then $c$ is

A)4
B)5
C)6
D)7

Attempt:
One way I know of to solve such a problem is by using the multinomial theorem but I am reluctant to use it here. It would take a lot of time to write down all the combinations and perform the calculations if I use multinomial theorem. Since this is an exam problem, I feel there exists a much better method.

Any help is appreciated. Thanks!

Hi Pranav,:)

I don't know if my way of attacking this problem looks easy (or the other way round) to you, but I think I will post my solution here...

$\displaystyle (3x^2+2x+c)^{12}=(3x^2+(2x+c))^{12}=\sum_{n=0}^{12} {12\choose n} (3x^2)^n(2x+c)^{12-n}$

Now, in order to collect all coefficients of $x^{19}$, we only need to consider the cases when $n=7,\,8,\,9$ (we could tell these are the only cases we need to take care if we think of the term $(3x^2)^n$:

[TABLE="class: grid, width: 70%"]
[TR]
[TD="width: 15%"]$n$[/TD]
[TD]The term associates with $x^{19}$[/TD]
[/TR]
[TR]
[TD]$n=7$[/TD]
[TD]${12\choose 7}\cdot 3^7\cdot{5\choose 0}\cdot2^{5} \cdot c^0\cdot x^{19}=25344(3^7)x^{19}$[/TD]
[/TR]
[TR]
[TD]$n=8$[/TD]
[TD]${12\choose 8}\cdot 3^8\cdot{4\choose 1}\cdot2^{3} \cdot c^1\cdot x^{19}=47520(3^7)cx^{19}$[/TD]
[/TR]
[TR]
[TD]$n=9$[/TD]
[TD]${12\choose 9}\cdot 3^9\cdot{3\choose 2}\cdot2^{1} \cdot c^2\cdot x^{19}=11880(3^7)c^2x^{19}$[/TD]
[/TR]
[/TABLE]

Now, in order to collect all coefficients of $x^{5}$, we only need to consider the cases when $n=0,\,1,\,2$ (similarly, we could tell these are the only cases if we think of the term $(3x^2)^n$:

[TABLE="class: grid, width: 70%"]
[TR]
[TD="width: 15%"]$n$[/TD]
[TD]The term associates with $x^{5}$[/TD]
[/TR]
[TR]
[TD]$n=0$[/TD]
[TD]${12\choose 0} \cdot{12\choose 7}\cdot 2^{5} \cdot c^7\cdot x^{5}=25344(c^7)x^{5}$[/TD]
[/TR]
[TR]
[TD]$n=1$[/TD]
[TD]${12\choose 1}\cdot 3^1\cdot{11\choose 8}\cdot2^{3} \cdot c^8\cdot x^{5}=47520(c^8)x^{5}$[/TD]
[/TR]
[TR]
[TD]$n=2$[/TD]
[TD]${12\choose 2}\cdot 3^2\cdot{10\choose 9}\cdot2^{1} \cdot c^9\cdot x^{5}=11880(c^9)x^{5}$[/TD]
[/TR]
[/TABLE]

Since we're told $\dfrac{A_{19}}{A_5}=\dfrac{1}{2^7}$ and from our working, we have

$\begin{align*}\dfrac{A_{19}}{A_5}&=\dfrac{25344(3^7)+47520(3^7)c+11880(3^7)c^2}{25344(c^7)+47520(c^8)+11880(c^9)}\\&=\dfrac{(3^7)(25344+47520c+11880c^2)}{(c^7)(25344+47520c+11880c^2}\\&=\dfrac{3^7}{c^7}\end{align*}$

In other words, we obtain $\dfrac{A_{19}}{A_5}=\dfrac{1}{2^7}=\dfrac{3^7}{c^7}$ or simply, $c=6$.

 

[Opalg]

Problem:
Let $\displaystyle (3x^2+2x+c)^{12}=\sum_{r=0}^{24} A_rx^r$ and $\dfrac{A_{19}}{A_5}=\dfrac{1}{2^7}$, then $c$ is

A)4
B)5
C)6
D)7

Attempt:
One way I know of to solve such a problem is by using the multinomial theorem but I am reluctant to use it here. It would take a lot of time to write down all the combinations and perform the calculations if I use multinomial theorem. Since this is an exam problem, I feel there exists a much better method.

Any help is appreciated. Thanks!

Here is a very intuitive and incomplete way of thinking about this problem.

In the expansion of $(3x^2+2x+c)^{12}$, each term will be a multinomial multiple of $(3x^2)^\alpha(2x)^\beta c^\gamma$, where $\alpha + \beta + \gamma = 12$. For a term of degree $5$, you need $2\alpha + \beta =5$, and for a term of degree $19$, you need $2\alpha + \beta =19$. For each term of degree $5$, if you decrease $\gamma$ by $7$ and increase $\alpha$ by $7$, then you will get a term of degree $19$, and every term of degree $19$ arises in this way. But replacing a $\gamma$ by an $\alpha$ involves replacing a $c$ by a $3$. It follows that $A_{19} = \Bigl(\dfrac3c\Bigr)^{\!7}A_5.$ If this is equal to $\Bigl(\dfrac12\Bigr)^{\!7}$ then clearly $c=6$.

I suspect that the "much better method" that you are looking for must somehow run along those lines. I am not at all convinced by my argument as it stands (although as anemone has shown, it leads to the right answer). It relies on the fact that the degree of the polynomial is $5+19 = 24$, so that the multinomial coefficients in $A_5$ and $A_{19}$ are the same. And the fact that $19-5 = 14$, which is twice $7$, also seems to be significant. Maybe someone can rescue this approach and make it more convincing.

 

[Saitama]

Thanks anemone and Opalg for the helpful replies. :)

But the method presented by anemone is still very long and involves a lot of calculations. If there is really no other way, I don't think its worth spending time on this problem during the exam where you have 30 questions for each of Physics, Chemistry and Maths and all are equally difficult. (Thinking)

EDIT: Ok, I asked for the given solution (to my friend from whom I borrowed the question paper) and here it goes:

$$(ax^2+bx+c)^n=\sum_{r=0}^{2n} A_rx^r$$
Putting $x=\frac{c}{ax}$, we get
$$(ax^2+bx+c)^n=\sum_{r=0}^{2n} A_{2n-r}\left( \frac{c}{a}\right)^{n-r}x^r$$

Hence, $\frac{A_r}{A_{2n-r}}=\left(\frac{c}{a}\right)^{n-r} \forall \,\,\,0\leq r\leq 2n$.

Put $r=19$,$n=12$, $a=3$
$$\Rightarrow \left(\frac{c}{3}\right)^{-7}=2^7 \Rightarrow c=6$$

I must be an idiot but I don't get even the first step the solution did. How did they arrive at the expression after the substitution?

 

[I like Serena]

I must be an idiot but I don't get even the first step the solution did. How did they arrive at the expression after the substitution?

Hey Pranav!

I can follow the solution (but I can't tell you how to think of it ).

The substitution is executed as:
\begin{aligned}
x^{2n-r} &= x^{n-r}x^n \\
&= \left(\frac c{ax}\right)^{n-r}x^n \\
&= \left(\frac c{a}\right)^{n-r}x^{r-n}x^n \\
&= \left(\frac c{a}\right)^{n-r}x^r \\
\end{aligned}

 

[Saitama]

Hi ILS! Sorry for the delay in reply.

Hey Pranav!

I can follow the solution (but I can't tell you how to think of it ).

The substitution is executed as:
\begin{aligned}
x^{2n-r} &= x^{n-r}x^n \\
&= \left(\frac c{ax}\right)^{n-r}x^n \\
&= \left(\frac c{a}\right)^{n-r}x^{r-n}x^n \\
&= \left(\frac c{a}\right)^{n-r}x^r \\
\end{aligned}

This is what I tried:

$$\left(a\frac{c^2}{a^2x^2}+\frac{bc}{ax}+c\right)^n=\sum_{r=0}^{2n}A_r\left(\frac{c}{ax}\right)^r$$

$$\Rightarrow \left(\frac{c^2+bcx+acx^2}{ax^2}\right)=\sum_{r=0}^{2n} A_r \left(\frac{c}{a}\right)^r\frac{1}{x^r}$$

$$\Rightarrow \left(\frac{c}{a}\right)^n \frac{1}{x^{2n}} \left(ax^2+bx+c\right)^n = \sum_{r=0}^{2n}A_r\left(\frac{c}{a}\right)^r\frac{1}{x^r}$$

$$\Rightarrow \left(ax^2+bx+c\right)^n=\sum_{r=0}^{2n} A_r \left(\frac{c}{a}\right)^{r-n} x^{2n-r}$$

I don't see how the solution wrote $A_{2n-r}$ instead of $A_r$.

 

[Opalg]

This is what I tried:

$$\left(a\frac{c^2}{a^2x^2}+\frac{bc}{ax}+c\right)^n=\sum_{r=0}^{2n}A_r\left(\frac{c}{ax}\right)^r$$

$$\Rightarrow \left(\frac{c^2+bcx+acx^2}{ax^2}\right)=\sum_{r=0}^{2n} A_r \left(\frac{c}{a}\right)^r\frac{1}{x^r}$$

$$\Rightarrow \left(\frac{c}{a}\right)^n \frac{1}{x^{2n}} \left(ax^2+bx+c\right)^n = \sum_{r=0}^{2n}A_r\left(\frac{c}{a}\right)^r\frac{1}{x^r}$$

$$\Rightarrow \left(ax^2+bx+c\right)^n=\sum_{r=0}^{2n} A_r \left(\frac{c}{a}\right)^{r-n} x^{2n-r}$$

I don't see how the solution wrote $A_{2n-r}$ instead of $A_r$.

If you make the substitution $s = 2n-r$ in the sum $$\sum_{r=0}^{2n} A_r \left(\frac{c}{a}\right)^{r-n} x^{2n-r}$$, then it becomes $$\sum_{s=0}^{2n} A_{2n-s} \left(\frac{c}{a}\right)^{n-s} x^s.$$ Having done that, you can re-label the summation index as $r$ instead of $s$.

 

[I like Serena]

I don't see how the solution wrote $A_{2n-r}$ instead of $A_r$.

Wait. Wait.
Let's walk through the solution step by step.

$$(ax^2+bx+c)^n=\sum_{r=0}^{2n} A_rx^r$$

We can reverse the order of the summation to get:
$$\sum_{r=0}^{2n} A_rx^r = \sum_{r=0}^{2n} A_{2n-r} x^{2n-r} \qquad (1)$$

Putting $x=\frac{c}{ax}$, we get
$$(ax^2+bx+c)^n=\sum_{r=0}^{2n} A_{2n-r}\left( \frac{c}{a}\right)^{n-r}x^r$$

Forget the left hand side. We're not doing anything with that.

We apply the substitution I mentioned earlier to get:
$$\sum_{r=0}^{2n} A_{2n-r} x^{2n-r} = \sum_{r=0}^{2n} A_{2n-r} \left(\frac c a\right)^{n-r}x^r\qquad (2)$$

Combine (1) and (2) to find:
$$\sum_{r=0}^{2n} A_rx^r = \sum_{r=0}^{2n} A_{2n-r} \left(\frac c a\right)^{n-r}x^r$$

Since the powers of $x$ must match left and right, we can tell that for each $r$ we have:
$$A_r = A_{2n-r} \left(\frac c a\right)^{n-r}$$

Then we can rewrite this as:

Hence, $\frac{A_r}{A_{2n-r}}=\left(\frac{c}{a}\right)^{n-r} \forall \,\,\,0\leq r\leq 2n$.

Finally, we get:

Put $r=19$,$n=12$, $a=3$
$$\Rightarrow \left(\frac{c}{3}\right)^{-7}=2^7 \Rightarrow c=6$$

 

[Saitama]

If you make the substitution $s = 2n-r$ in the sum $$\sum_{r=0}^{2n} A_r \left(\frac{c}{a}\right)^{r-n} x^{2n-r}$$, then it becomes $$\sum_{s=0}^{2n} A_{2n-s} \left(\frac{c}{a}\right)^{n-s} x^s.$$ Having done that, you can re-label the summation index as $r$ instead of $s$.

Great! Thanks a lot Opalg! :D

It is very difficult to think of this solution during the exam.

Forget the left hand side. We're not doing anything with that.

We apply the substitution I mentioned earlier to get:
$$\sum_{r=0}^{2n} A_{2n-r} x^{2n-r} = \sum_{r=0}^{2n} A_{2n-r} \left(\frac c a\right)^{n-r}x^r\qquad (2)$$

Combine (1) and (2) to find:
$$\sum_{r=0}^{2n} A_rx^r = \sum_{r=0}^{2n} A_{2n-r} \left(\frac c a\right)^{n-r}x^r$$

Since the powers of $x$ must match left and right, we can tell that for each $r$ we have:
$$A_r = A_{2n-r} \left(\frac c a\right)^{n-r}$$

Very nicely explained, thanks a lot ILS! :)