Binomial Expansion

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carlos1017

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Question: Find the sum of the first 4 terms of the binomial expansion (1+2)1/3
I tried using the Binomial Theorem formula:
Screen Shot 2015-05-13 at 11.45.03 PM.png
But I'm not sure how to do factorial fractions (that is if its even possible). Maybe you're not suppose to use the Binomial Theorem?
Thanks!
 
carlos1017 said:
Question: Find the sum of the first 4 terms of the binomial expansion (1+2)1/3
I tried using the Binomial Theorem formula:
View attachment 5309
But I'm not sure how to do factorial fractions (that is if its even possible). Maybe you're not suppose to use the Binomial Theorem?
Thanks!
Click to expand...
Try using (1+x)​^n​​=1+nx+​2!​​n(n−1)​​x​^2​​+​3!​​n(n−1)(n−2)​​x^​3​ where x=2 and n=1/3
 
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carlos1017 said:
Question: Find the sum of the first 4 terms of the binomial expansion (1+2)1/3
I tried using the Binomial Theorem formula:
View attachment 5309
But I'm not sure how to do factorial fractions (that is if its even possible). Maybe you're not suppose to use the Binomial Theorem?
Thanks!
Click to expand...
Actually the sum is to infinity but if n is a positive integer, all terms after the nth one are zero. So, if we write out the binomial coefficient (and use a as a generic value instead of n) we have
\(\displaystyle \begin{pmatrix}a\\0\end{pmatrix}\, =\, 1\)
\(\displaystyle \begin{pmatrix}a\\k\end{pmatrix}\, =\, \frac{a\, (a-1)\, (a-2)\, ...\, (a-k+1)}{k!}\)
 
Jomo said:
Try using (1+x)​^n​​=1+nx+​2!​​n(n−1)​​x​^2​​+​3!​​n(n−1)(n−2)​​x^​3​ where x=2 and n=1/3
Click to expand...

Ok, I came up with 159/9, what is that value though? , is it the fourth term or the sum ?
 
Ishuda said:
Actually the sum is to infinity but if n is a positive integer, all terms after the nth one are zero. So, if we write out the binomial coefficient (and use a as a generic value instead of n) we have
\(\displaystyle \begin{pmatrix}a\\0\end{pmatrix}\, =\, 1\)
\(\displaystyle \begin{pmatrix}a\\k\end{pmatrix}\, =\, \frac{a\, (a-1)\, (a-2)\, ...\, (a-k+1)}{k!}\)
Click to expand...

Right aC0 is 1, but n in this problem is not a positive integer that's why i'm confused, is there a way to evaluate 1/3C5? (i'm not sure if that'll work either)
Also, n choose k is n!/k!(n-k)!
 
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I'm not really sure where Jomo came up with his sum, nor do I know what 159/9 signifies. However, I have an alternate solution, which may or may not be any easier for you. The difficulty is that the exponent is a fraction. The binomial theorem involves a sum from k = 0 to n. But here n is a fraction, and I don't believe the sigma operator takes fractional bounds. My solution is to rewrite the problem, such that the exponent is no longer a fraction, like so:

\(\displaystyle 3^{\frac{1}{3}}=\left(3^{\frac{1}{6}}\right)^2\)

And then you can apply the binomial theorem, with n = 2, a = 1, and b = 3^(1/6) - 1, and everything will work out fine. I believe you can use any positive values of a and b as long as they add up to the sixth root of 3, but I choose a = 1 for simplicity's sake. Do note, though, that neither a nor b can be zero.
 
ksdhart said:
I'm not really sure where Jomo came up with his sum, nor do I know what 159/9 signifies. However, I have an alternate solution, which may or may not be any easier for you. The difficulty is that the exponent is a fraction. The binomial theorem involves a sum from k = 0 to n. But here n is a fraction, and I don't believe the sigma operator takes fractional bounds. My solution is to rewrite the problem, such that the exponent is no longer a fraction, like so:

\(\displaystyle 3^{\frac{1}{3}}=\left(3^{\frac{1}{6}}\right)^2\)

And then you can apply the binomial theorem, with n = 2, a = 1, and b = 3^(1/6) - 1, and everything will work out fine. I believe you can use any positive values of a and b as long as they add up to the sixth root of 3, but I choose a = 1 for simplicity's sake. Do note, though, that neither a nor b can be zero.
Click to expand...

That's a good solution, I tried to come up with the first 4 terms but the 4th term would have 2C3, is it possible to evaluate 2!/3!(2-3)! ? (-1!)
 
Hmm. I didn't see that part of your problem at first. Since the problem does specify you need to have four terms, then I'm at a loss as to how to proceed. I was able to work the problem out and arrive at the correct answer, but my sum has only three terms. Perhaps someone else can help you more.

You are definitely correct about not being able to evalulate (-1)! The factorial function handles only positive values. Fractional values can be solved, but I'd leave it to a computer because you'd need to evaluate an improper integral. And you likely don't even know an integral is, much less how to evaluate one ;)

EDIT: So I just thought of this. What if you used n = 3, a = 1, and b = 3^(1/9)-1. Because 3^(1/9) cubed is 3^(1/3). That way you would be able to have 4 terms, since the final term would be 3C3. I haven't actually worked this out myself, but it seems plausible.
 
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carlos1017 said:
Right aC0 is 1, but n in this problem is not a positive integer that's why i'm confused, is there a way to evaluate 1/3C5? (i'm not sure if that'll work either)
Also, n choose k is n!/k!(n-k)!
Click to expand...

Indeed n choose k is n!/[k!(n-k)!]EDIT! Given that k is an integer for nCk, then
\(\displaystyle \begin{pmatrix}n\\k\end{pmatrix}\, =\, \frac{n!}{k!\, (n-k)!}\)
\(\displaystyle =\, \frac{n\, (n-1)\, (n-2)\, ...\, (n-k+1)\, (n-k)!}{k!\, (n-k)!}\)
\(\displaystyle =\, \frac{n\, (n-1)\, (n-2)\, ...\, (n-k+1)}{k!}\)

So, for example
\(\displaystyle \begin{pmatrix}\frac{1}{3}\\3\end{pmatrix}\, =\, \frac{\frac{1}{3}\, (\frac{1}{3}-1)\, (\frac{1}{3}-2)}{3!}\)
\(\displaystyle \begin{pmatrix}\frac{1}{3}\\4\end{pmatrix}\, =\, \frac{\frac{1}{3}\, (\frac{1}{3}-1)\, (\frac{1}{3}-2)\, (\frac{1}{3}-3)}{4!}\)
 
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carlos1017 said:
Question: Find the sum of the first 4 terms of the binomial expansion (1+2)1/3
I tried using the Binomial Theorem formula:
View attachment 5309
But I'm not sure how to do factorial fractions (that is if its even possible). Maybe you're not suppose to use the Binomial Theorem?
Thanks!
Click to expand...
Hi, I am not sure why ksdhart is prolonging things here. You were given the formula to find the 1st four terms which you used. I did not check to see if it comes out to 159/9 or not but again the formula is correct.
 
carlos1017 said:
Question: Find the sum of the first 4 terms of the binomial expansion (1+2)1/3
I tried using the Binomial Theorem formula:
View attachment 5309
But I'm not sure how to do factorial fractions (that is if its even possible). Maybe you're not suppose to use the Binomial Theorem?
Thanks!
Click to expand...

One last thing; when you use the [extended] binomial theorem for problems of this kind, you should note that a must be larger than b and the upper limit for the sum is infinity, see
http://en.wikipedia.org/wiki/Binomial_theorem#Generalisations
for example.
 
Jomo said:
Hi, I am not sure why ksdhart is prolonging things here. You were given the formula to find the 1st four terms which you used. I did not check to see if it comes out to 159/9 or not but again the formula is correct.
Click to expand...

I apologize if I come off as rude, but I offered an alternate solution because your formula does not produce the correct answer. It indeed has four terms, but the answer has to come out to 3^(1/3). Your sum evaluates out to 159/9 = 53/3, as Carlos noted. That's 17 and 2/3, which is not evenclose to the third root of 3 (~1.4422).
 
ksdhart said:
Hmm. I didn't see that part of your problem at first. Since the problem does specify you need to have four terms, then I'm at a loss as to how to proceed. I was able to work the problem out and arrive at the correct answer, but my sum has only three terms. Perhaps someone else can help you more.

You are definitely correct about not being able to evalulate (-1)! The factorial function handles only positive values. Fractional values can be solved, but I'd leave it to a computer because you'd need to evaluate an improper integral. And you likely don't even know an integral is, much less how to evaluate one ;)

EDIT: So I just thought of this. What if you used n = 3, a = 1, and b = 3^(1/9)-1. Because 3^(1/9) cubed is 3^(1/3). That way you would be able to have 4 terms, since the final term would be 3C3. I haven't actually worked this out myself, but it seems plausible.
Click to expand...

Yeah thats what i also thought during school, the it would end with 3 choose 3 , just like a 3 degree binomial. Indeed its plausible, but it would be pretty hard to evaluate it in 4 minutes or less :v.
Anyways thank you so much for helping :)
 
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