MHB Coefficient of trinomial expansion

AI Thread Summary
The discussion focuses on finding the coefficients of \(x^4\) and \(x^{39}\) in the expansion of \((1+x+x^2)^{20}\). The coefficient of \(x^4\) is calculated using combinatorial methods, resulting in a total of 8455 after correcting initial miscalculations. For \(x^{39}\), the coefficient is determined to be \(20\), derived from the binomial coefficient \({20 \choose 19}\). Participants also explore a general approach for finding coefficients in trinomial expansions, suggesting a method involving completing the square. The conversation highlights the complexity and potential for errors in manual calculations, emphasizing the utility of computational tools for verification.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Hi,
I want to find the coefficient of $\displaystyle x^4 $ and $\displaystyle x^{39} $ of $\displaystyle (1+x+x^2)^{20} $

I use the binomial theorem twice and come out with the following argument:
$\displaystyle (1+x+2x^2)^{20}= [(1+x)+(2x^2)]^{20}$

I see that
$\displaystyle T(r+1)={20 \choose r} (1+x)^{20-r}(2x^2)^r $

$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.(1+x)^{20-r} $

$\displaystyle T(r+1)={20 \choose r} .2^r. x^{2r}.{20-r \choose k} (1)^{20-r-k}. (x)^{k}$

$\displaystyle T(r+1)={20 \choose r} {20-r \choose k} .2^r. x^{2r+k}$

Now, I notice that $\displaystyle 20-r \ge k$, and, of course, $\displaystyle 0\le r,k \le 20 $.

To obtain the coefficient of x^4, we need 2r+k=4
If k=0, r=2. (Case 1)...this gives the coefficient of 760.
If k=1, then r is not an integer.
If k=2, r=1 (Case 2)...this gives the coefficient of 6840.
If k=4, r=0 (Case 3)...this gives the coefficient of 4845.

Hence, the coefficient of x^4 is (760+6840+4845=12445).

And I do the same for x^39. It turns out that there is only one case to generate the term x^39, i.e. when k=1, r=19. Thus the coefficient of x^39 is 20(2^19).

All this is only common sense and it certainly is not an elegant method to find for coefficient of any specific term. I wish to ask if there exists a general formula to find the coefficient of trinomial expansion of the type (a+bx+cx^2)?

Thanks.
 
Last edited:
Mathematics news on Phys.org
Three words: completing the square.

anemone said:
All this is only common sense and it certainly is not an elegant method to find for coefficient of any specific term. I wish to ask if there exists a general formula to find the coefficient of trinomial expansion of the type (a+bx+cx^2)?

Thanks.
Let $h = -\frac{b}{2a}$ and $j = c-\frac{b^2}{4a}$, then $ax^2+bx+c = a(x-h)^2+j$. We have:$$\begin{aligned} \displaystyle (ax^2+bx+c)^n & = j^n \left(\frac{a}{j}(x-h)^2+1\right)^n \\& = j^n \sum_{0 \le k \le n}\binom{n}{k} \frac{a^k}{j^k} (x-h)^{2k} \\& = \sum_{0 \le k \le n}\binom{n}{k} a^k j^{n-k} h^{2k} (1-\frac{1}{h}~x)^{2k} \\& = \sum_{0 \le k \le n}~ \sum_{0 \le r \le 2k}\binom{n}{k}\binom{2k}{r}a^k j^{n-k}h^{2k-r}(-1)^rx^r.\end{aligned}$$

---------- Post added at 12:45 PM ---------- Previous post was at 11:15 AM ----------

For example, using the same technique we find the trinomial:

$\displaystyle (1+x+x^2)^n = \sum_{0 \le k \le n}~~\sum_{0 \le r \le 2k}\binom{n}{k}\binom{2k}{r} 4^{k-n}3^{n-k}2^{r-2k}x^r$

So the coefficient of $x^r$ is $\displaystyle \sum_{0 \le k \le n}\binom{n}{k}\binom{2k}{r} 4^{k-n}3^{n-k}2^{r-2k}.$
 
Last edited:
Thanks, Sherlock. That works beautifully!But in your example below, if I want to determine the coefficient of x^8, what would the k value be?
(P.S. I'm sorry for being so dumb to ask for this one but I really have no idea...(Sadface)...)

$\displaystyle (1+x+x^2)^{20} = \sum_{0 \le k \le 20}~~\sum_{0 \le r \le 2k}\binom{20}{k}\binom{2k}{r} 4^{k-20}3^{20-k}2^{r-2k}x^r$

So the coefficient of $x^8$ is $\displaystyle \sum_{0 \le k \le 20}\binom{20}{k}\binom{2k}{r} 4^{k-20}3^{20-k}2^{8-2k}.$
 
anemone said:
Hi,
I want to find the coefficient of $\displaystyle x^4 $ and $\displaystyle x^{39} $ of $\displaystyle (1+x+x^2)^{20} $
[snip..]

And I do the same for x^39. It turns out that there is only one case to generate the term x^39, i.e. when k=1, r=19. Thus the coefficient of x^39 is 20(2^19).

I'm not sure what you intend for the coefficient of \(x^{39}\) here, but it is obviously \(20\).

Also with a bit of ingenuity one can work out the coefficient of \(x^4\) by considering how to get \(x^4\) from multiplying out the \(20\) brackets containing \( (1+x+x^2) \)

CB
 
Last edited:
anemone said:
Hi,
I want to find the coefficient of $\displaystyle x^4 $ and $\displaystyle x^{39} $ of $\displaystyle (1+x+x^2)^{20} $
For the coefficient of $x^4 $, you could write $1+x+x^2 = \dfrac{1-x^3}{1-x}$. Then $$(1+x+x^2)^{20} = (1-x^3)^{20}(1-x)^{-20} = \bigl(1-20x^3+\ldots)\bigr)\bigl(1+20x+210x^2+1540x^3+7084x^4+\ldots\bigr),$$ from which you can pick out the coefficient of $x^4$ as $7084-400 = 6684.$
 
opalg said:
for the coefficient of $x^4 $, you could write $1+x+x^2 = \dfrac{1-x^3}{1-x}$. Then $$(1+x+x^2)^{20} = (1-x^3)^{20}(1-x)^{-20} = \bigl(1-20x^3+\ldots)\bigr)\bigl(1+20x+210x^2+1540x^3+7084x^4+\ldots\bigr),$$ from which you can pick out the coefficient of $x^4$ as $7084-400 = 6684.$
8455

cb
 
Last edited:
CaptainBlack said:
I'm not sure what you intend for the coefficient of \(x^{39}\) here, but it is obviously \(20\).

Hmm...sometimes, the word 'obviously' in the context of math is intimidating.:cool:
But now that I can see and yes, it's quite obvious because $x^{39}$ is the second last term from the expansion and its coefficient should be $ \displaystyle {20\choose19} $

CaptainBlack said:
Also with a bit of ingenuity one can work out the coefficient of \(x^4\) by considering how to get \(x^4\) from multiplying out the \(20\) brackets containing \( (1+x+x^2) \)

OK.

Thanks.

---------- Post added at 05:41 AM ---------- Previous post was at 05:36 AM ----------

Opalg said:
For the coefficient of $x^4 $, you could write $1+x+x^2 = \dfrac{1-x^3}{1-x}$. Then $$(1+x+x^2)^{20} = (1-x^3)^{20}(1-x)^{-20} = \bigl(1-20x^3+\ldots)\bigr)\bigl(1+20x+210x^2+1540x^3+7084x^4+\ldots\bigr),$$ from which you can pick out the coefficient of $x^4$ as $7084-400 = 6684.$

Thanks, Opalg!
You have given me a great idea and I really appreciate that.
By the way, I noticed that the coefficient of $x^4$ should be ${23\choose 4}=8855$, :).
 
anemone said:
Hmm...sometimes, the word 'obviously' in the context of math is intimidating.:cool:
But now that I can see and yes, it's quite obvious because $x^{39}$ is the second last term from the expansion and its coefficient should be $ \displaystyle {20\choose19} $
OK.

Thanks.

---------- Post added at 05:41 AM ---------- Previous post was at 05:36 AM ----------



Thanks, Opalg!
You have given me a great idea and I really appreciate that.
By the way, I noticed that the coefficient of $x^4$ should be ${23\choose 4}=8855$, :).

Typo? 8455

CB
 
CaptainBlack said:
Typo? 8455

CB

Ah! Typo! It should be 8855-400=8455.
 
  • #10
anemone said:
Ah! Typo! It should be 8855-400=8455.
Yes, I'm not sure how I got 7084 instead of 8855. That comes from trying to do arithmetic on paper instead of reaching for a calculator.
 
  • #11
Opalg said:
Yes, I'm not sure how I got 7084 instead of 8855. That comes from trying to do arithmetic on paper instead of reaching for a calculator.

After having done the calculation by hand, I reached for something far more powerful than a calculator to check (expand (1+x+x^2)^20 in maxima or in Wolfram Alpha - at least once you get past the offer of a free trial of Alpha-pro).

CB
 
Back
Top