# I'm having difficulty expressing a binomial expansion as a sum

#### Robin64

Problem Statement
Given a function, f(x)=(8-3x^2)^-(⅓), find the 4 terms of that function's binomial expansion and then represent that expansion as a sum
Relevant Equations
f(x)=(8-3x^2)^-(⅓), (1+x)^n=1+nx+n(n-1)(x^2)/2!+n(n-1)(n-2)(x^3)/3!
I found the first 4 terms of the series: ½-(1/16)x^2+(1/64)x^4-(7/1536)x^6.

I cannot however simplify this to a sum. the 7 in the numerator of the last term of the above expansion is the sticking point.

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#### Mark44

Mentor
I found the first 4 terms of the series: ½-(1/16)x^2+(1/64)x^4-(7/1536)x^6.

I cannot however simplify this to a sum. the 7 in the numerator of the last term of the above expansion is the sticking point.
You have a formula for $(1 + x)^n$. Make life easier on yourself by writing $(8 - 3x^2)^{-1/3}$ as $8^{-1/3}(1 - \frac 3 8 x^2)^{-1/3}$ and then replacing $-\frac 3 8 x^2$ with, say u.

Then you're working with $(1 + u)^{-1/3}$, for which you know a formula, and presumably can come up with the general term.

As a tip, leave the numbers unmultiplied in those fractions. It's much harder to determine a pattern with 1/64 and 7/1536 and so on.

#### Ray Vickson

Science Advisor
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The general binomial coefficient of $x^k$ in the expansion $(1+x)^{\lambda}$ is
$${\lambda \choose k} \equiv \frac{\lambda (\lambda -1) \cdots (\lambda-k+1)}{k!}, \: k = 0,1, 2, 3, \ldots$$

Newton proved this as an application of his calculus of fluxions.

So
$$(1-u)^{-1/3} = 1 +(-u) {-1/3 \choose 1}+(-u)^2 {-1/3 \choose 2} + \cdots$$

Note that
$${-1/3 \choose 1} = -1/3$$
$${ -1/3 \choose 2} = \frac{(-1/3)(-1/3-1)}{2!} = (-1)^2 \frac{(1/3)(4/3)}{2} = \frac{4}{18}$$ etc.

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#### Robin64

In my work, I did re-write the function as (½)(1-3x^2)^-⅓. The problem is the pesky 7. I don't see a pattern that gives me a closed form solution. Here's where I'm at:

#### Ray Vickson

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In my work, I did re-write the function as (½)(1-3x^2)^-⅓. The problem is the pesky 7. I don't see a pattern that gives me a closed form solution. Here's where I'm at:View attachment 241661
The binomial coefficient can be written as
$${-1/3 \choose k} =(-1)^k\, \frac{\prod_{j=0}^{k-1} (1+3 j)}{3^k k!}.$$
I don't think it can be expressed much more compactly than that.

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#### Robin64

Thanks. I ended up expressing the sum as follows (I don't know the right code to display this properly, so I just took a screenshot of my solution as represented in Maple Document):

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#### Ray Vickson

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Thanks. I ended up expressing the sum as follows (I don't know the right code to display this properly, so I just took a screenshot of my solution as represented in Maple Document):
Right. But the binomial coefficient can be re -written in more explicit form; see \$5 above.

#### Mark44

Mentor
Notice that the numerator of those fractions are -1, then -1 - 4, then -1 - 4 - 7, ... Do you see a pattern? What will the next numerator be?

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